ADVANCED GEOMETRY
FOE HIGH SGHOOLS
SYNTHETIC ANALYTICAL
M3DOUGALL.
REVISED EDITION
Digitized by the Internet Archive
in 2009 with funding from
Ontario Council of University Libraries
http://www.archive.Org/details/1 91 9advancedgeom00mcdo
ADVANCED GEOMETRY
FOR
HIGH SCHOOLS
SYNTHETIC AND ANALYTICAL
PART I.— SYNTHETIC GEOMETRY
PART II.— ANAL YTICAL GEOMETR Y
A. H. McDOUGALL, B.A., LL.D.
Principal Ottawa Collegiate Institute
SECOND EDITION
REVISED AXD AMPLIFIED
TORONTO
THE COPP CLARK COMPANY, LIMITED
Copyright, Canada, 1919, by The Copp Clark Company, Limited,
Toronto, Ontario
PREFACE TO THE SECOND EDITION
In this edition changes have been made in Part I. on the
suggestion of teachers who have kindly pointed out difficulties
experienced by their pupils in using the original edition. The
exercises have been rearranged and, in general, divided into
two sections. Those marked (a) are intended for a first
reading, and are sufficient for candidates for Entrance to
the Faculties of Education, while the sections marked (b)
are more suitable for candidates for honours and scholarships.
Demonstrations have been rewritten, diagrams added,
and notes given suggesting constructions or proofs. Some
typical solutions illustrate methods to be used in maxima and
minima.
Tn many cases, e.g., loci, the difficulties have been di
minished. Certain exercises in the First Edition were stated
as problems. Under the given conditions the students were
asked to find the required solution. These exercises are here
changed to theorems. With the same data the conclusion is
stated and the proof only is to be discovered.
In both parts considerable additions suitable for honour
candidates have been made to the miscellaneous exercises.
Acknowledgments for valuable assistance received from
him are due to Mr. I. T. Norris, Mathematical Master of the
Ottawa Collegiate Institute, and to the teachers who have
given an encouraging approval of the book as well as assist
ance in making it more useful.
Ottawa, February, 1919.
CONTENTS
Chapter I
l'AQB
Theorems of Menelaus and Ceva 1
The NinePoint Circle 11
Simpson's Line 13
Areas op Rectangles 18
Radical Axis 23
Chapter II
Medial Section 30
Miscellaneous Theorems 37
Similar and Similarly Situated Polygons ... 39
Chapter III
Harmonic Ranges and Pencils 43
The Complete Quadrilateral 49
Poles and Polars 50
Maxima and Minima 60
Miscellaneous Exercises
Loci 68
Theorems • '3
Problems • °*
Part I.— SYNTHETIC GEOMETRY
SYNTHETIC GEOMETRY
CHAPTER I
Theorems of Menelaus and Ceva
1. Menelaus' Theorem:— If a transversal cut the
sides, or the sides produced, of a triangle, the
product of one set of alternate segments taken in
circular order is equal to the product of the other set.
(Note. — The transversal must cut two sides and the third
side produced, or cut all three produced.}
C D^
Fiq. 1. Fia. 2.
A transversal cuts the sides BC, CA, AB of the A
ABC in the points D, E, F respectively, then
AF . BD . CE = FB . DC . EA.
Draw AX, BY, CZ j_ to the transversal.
From similar As : — ^Fx, P F Y
AF = AX
FB BY'
sss
a CE CZ
cEz.,*FX and ^A = ^
By multiplication
A> x ^X 21 1 K£*H?
FB X DC X EA  i '
and /. AF . BD . CE = FB . DC . EA.
(Historical Note.— Menelaus, a Greek, lived in Alexandria, Egypt, about 98 A.D.)
1
2 SYNTHETIC GEOMETRY
2. Converse of Menelaus' Theorem :— If, in A abc
on two of the sides bc, CA, ab and on the third
produced, or if on all three produced, points D, E, F
respectively be taken so that AF . BD . CE= FB. DC. EA,
the points D, E, F are collinear.
Join EF, and produce EF to cut BC at Q.
V FEG is a st. line,
.'. AF . BG . CE = FB . GC . EA. (§ I.)
But AF . BD . CE = FB . DC . EA, by hypothesis.
,. j. BG
dividing, — —
&' BD
BG
GC.
DC'
BD
DC*
or, by alternation, QC QC
/. G coincides with D. (O.H.S. Geometry, § 121.)
.". D. E, F are collinear.
THEOREMS OF MENELAUS AND CEVA 6
3. Ceva's Theorem:— If from the vertices of a
triangle concurrent straight lines be drawn to cut
the opposite sides, the product of one set of alternate
segments taken in circular order is equal to the
product of the other set.
(Note. — D, E and F must be on the three sides, or on
one side and on the other two produced.)
AO, BO, CO drawn from the vertices of A ABC
cut BC, CA, AB, at D, E, F respectively, then
AF . BD . CE = FB . DC . EA.
FOC is a transversal of A ABD,
.. AF . BC . DO = FB . CD . OA. (§ 1.)
BOE is a transversal of A ADC,
/. AO . DB. CE = OD . BC . EA.
By multiplication, and division by DO, OA and BC,
AF . BD . CE = FB . DC . EA.
(For another proof of this theorem see O.H. S.
Geometry, § 122, Exercises 12 and 14.)
(Historical Not*.— Giovanni Ceva, an Italian engineer, died in 1734 A.D.)
4 SYNTHETIC GEOMETRY
4. Converse of Ceva's Theorem :— If, in A abc,
on the three sides BC, CA, AB, or if on one of these
sides and on the other two produced, points D, E, F
respectively be taken so that AF . bd . CE = FB . DC . EA,
the lines AD, be, cf are concurrent.
Fia. 7. Fig. 8.
Draw BE, CF and let them cut at O. Join AO and
let it cut BC at G.
•/ AG, BE, CF are concurrent,
.'. AF . BG . CE = FB . GC . EA, (§ 3.)
But AF . BD . CE  FB . DC . EA, by hypothesis.
,. .,. BG GC.
.*. , dividing, =r= = =—? ,
"' o> BD DC
Dr, by alternation, ^ = ^.
.*. G coincides with D.
.*. AD, BE, CF are concurrent.
THEOREMS OF MENELAUS AND CEVA 5
(J. The perpendiculars from the vertices of a
triangle to the opposite sides are concurrent.
c B
In A ABC, draw AX ± BC, BY ± CA, CZ J. AB.
To prove that AX, BY, CZ are concurrent.
A AZC III A AYB,
AZ _ CA
YA ~ AB'
tii^ AB
arl ^' ZB = BC ;
^Y _ BC
XC CA"
and
6,'sfiZ*
%
AZ
YA
BX
ZB
CY
XC
CA
AB
AB BC _ ,
BC " CA
AZ . BX . CY = ZB . XC . YA
:., by § 4, AX, BY, CZ are concurrent.
(@/>The point where the J_s from the vertices of a
A to the opposite sides intersect is called the ortho
centre of the A. The A formed by joining the feet
of these J_s, X, Y, Z in Figures 9 or 10, is called the
orthocentric, or pedal, A.
6 SYNTHETIC GEOMETRY
7.— Exercises
(a)
1. Show, from the converse of Ceva's Theorem, that the
medians of a A are concurrent.
2. In A ABC, the bisectors of the L s A, B, C cut BC,
be
CA, AB at D, E, F respectively. Show that AF = r 
a + b
If a = 25, b = 35, c = 20, show that AF . BD . CE =
2062 T Vr
3 Show, from the converse of Ceva's Theorem, that the
bisectors of the /.s of a A are concurrent.
4. In A ABC, the bisector of the interior L at A and
of the exterior Zs at B, C cut BC, CA, AB at D, E, F
be
respectively. Show that AF = r
If a = 44, b = 33, c = 22, show that AF . BD . CE =
76665 .
>5 Show, from the converse of Ceva's Theorem, that the
bisector of the Z at one vertex of a A and the bisectors
of the exterior Z_s at the other two vertices are concurrent.
6. In A ABC, AX, BY, CZ the ±a to BC, CA, AB
intersect at O. Show that : —
(a) rect. AO . OX = rect. BO . OY = rect. CO . OZ ;
(b) rect. AB . AZ = rect. AO . AX = rect. AC . AY;
\c) if AX meet the circumscribed circle of A ABC at K,
OX = XK ;
\d) ZAYZ = ZB= ZAOZ;
(e) As AYZ, BZX, CXY, ABC are similar;
{/) AX, BY, CZ bisect the As of the pedal A XYZ;
(g) of the four points A, B, C, O, each is the orthocentre
of the A of which the other three points are the vertices;
EXERCISES 7
(h) if a A LMN be formed by drawing through A, B, C
lines MN, NL, LM II BC, CA, AB respectively, O is the
circumscribed centre of A LMN;
(i) if S be the centre of the circumscribed circle of A
ABC, AS, BS, CS are respectively J_ YZ, ZX, XY the sides
of the orthocentric A.
7. In A ABC, the inscribed circle touches BC, CA, AB
at D, E, F respectively, and s = the semiperimeter. Show
that AF = s  a.
If a = 43, b = 31, c = 26, show that AF . BD . CE = 3192.
X8. The st. lines joining the vertices of a A to the points
of contact of the opposite sides with the inscribed circle
are concurrent.
i). In A ABC, an escribed circle touches BC at D and
AB, AC, produced at F, E respectively. Show that FB =
s  c.
If a = 40, 6 = 30, c = 50, show that AF . BD.CE= 18000.
*>10. The st. lines joining the vertices of a A to the
points of contact of the opposite sides with any one of the
escribed circles are concurrent.
11. O is a point within the A ABC and AO, BO, CO
produced cut BC, CA, AB at D, E, F respectively. The
circle through D, E, F cuts BC, CA, AB again at P, Q, R.
Show that AP, BQ, CR are concurrent.
(*)
<*12. The bisectors of L s B, C of A ABC cut CA, AB at
E, F respectively. FE, BC produced meet at D. Prove
that AD bisects the exterior Z. at A.
*13. The points where the bisectors of the exterior /s at
A, B, C of A ABC meet BC, CA, AB respectively are
collinear.
8
SYNTHETIC GEOMETRY
14. AB, CD, EF are three ) st. lines. AC, BD meet at
N; CE, DF at L; EA, FB at M. Prove that L, M, N are
collinear.
15. (a) If two As are so situated that the st. lines
joining their vertices in pairs are concurrent, the inter
sections of pairs of corresponding sides are collinear. —
Desargues' Theorem.
(Note. — ABC, abc the two As; Aa, Bb, Cc meeting at OJ
BC, ic at L; CA, ca at M ; AB, ab at N. Using the As
OBC, OCA, OAB and the respective transversals bcL, acM,
a&N prove that AN . BL . CM = NB . LC . MA.)
(b) State and prove the converse of (a).
(Note.— BC, JcmeetatL;
CA, ca at M ; AB, ab at N
and L, M, N are collinear.
Produce Act, Bb to meet at
O. To show that Cc passes
through O. A«M, B&L
are As having AB, ab, ML
concurrent at N ; corres
ponding sides aM, bLmeet
at c; MA, LB at C; Aa,
Bb at O. .*., by (a), c,
C, O are collinear, i.e., Aa,
N Bb, Cc are concurrent.)
Fig. 11.
(Historical Notb :— Girard Desargues (15931662) was an architect and engineer of
Lyons, France.)
16. The inscribed circle of A ABC touches the sides
BC, CA, AB at D, E, F respectively; EF, FD, DE, produced
meet BC, CA, AB respectively at L, M, N. Show that
L, M, N are collinear.
(Note.— Use Ex. 8 and Ex. 15 (a).)
EXERCISES 9
17. Tangents to the circumcircle at A, B, C, meet BC,
CA, AB respectively in collinear points.
(Note.— Use Ex. 8 and Ex. 15 (a).)
18. Given the base and vertical Z. of a A, find the locus
of its orthocentre.
19. If the base BC and vertical L A of a A ABC be
given, and the base be trisected at D, E, the locus of the
centroid is an arc containing an L equal to L A, and
having DE as its chord.
20. ABC is a A, XYZ its pedal A. Show that the
respective intersections of BC, CA, AB with YZ, ZX, XY
are collinear.
21. "Where is the orthocentre of a rt.Zd A 1 ?
22. If one escribed circle of A ABC touch AC at F and
BA produced at G, and another escribed circle touch AB
at H and CA produced at K, FH, KG produced cut BC
produced in points equidistant from the middle point of BC.
(Note. — Use § 1, taking the two transversals FH, KG of
A ABC and multiplying the results.)
23. If O is the orthocentre, S the circumcentre and I
the centre of the inscribed circle of A ABC, prove that
IA bisects L OAS.
10
SYNTHETIC GEOMETRY
8. The distance from each vertex of a triangle
to the orthocentre is twice the perpendicular from
the circumcentre to the side opposite that vertex.
O is the orthocentre, S the
circumcentre of A ABC; SD
is ± BC.
To show that AO = twice
SD.
Draw the diameter CSE,
join BE, EA.
Ls EBC, EAC being Ls in
fig. 12. semicircles are rt. Ls.
;. EB ! AO and EA II BY.
.*. EAOB is a  gm, and EB = AO.
But V S, D are middle points of EC, BC,
.*. EB = twice SD,
and /. AO = twice SD.
9. ABC is a A having AX _L
BC, AD a median, O the ortho
centre, and S the circumscribed
centre. Show that OS cuts AD
at the centroid G. (Use § 8.)
Show also that G is a point
of trisection in SO.
A general enunciation of these
results maybe given as follows: —
The Orthocentre, Centroid, and the centre of the
Circumscribed Circle of a A are in the same st
line, and the Centroid is a point of trisection in the
St. line joining the other two.
the ninepoint circle
The NinePoint Circle
11
10. The three middle points of the sides of a
triangle, the three projections of the vertices on the
opposite sides, and the three middle points of the
straight lines joining the vertices to the orthocentre
are all concyclic.
Let ABC be a A, AX, BY, CZ the J_s from A, B, C
to BC, CA, AB respectivel}', O the orthocentre, L, M, N
the middle points of AO, BO, CO respectively, D, E, F
the middle points of BC, CA, AB respectively.
It is required to show that the nine points, X, Y, Z,
L, M, N, D, E, F are concyclic.
Find S, the clrcumcentre. Draw SO, SD, SA. Biseci
SO at K. Draw KL.
V K, L are the middle points of SO, OA ;
.'. KL = half of SA;
and .*. the circle described with centre K and radius
equal to half that of the circumcircle passes through L.
12 SYNTHETIC GEOMETRY
Similarly this circle passes through M and N.
Draw DK.
f SD= LO, (§8)
In AS SKD, OKL^ SK= KO,
(zDSK=ZKOL, (SDULO);
.'. DK = KLand ZSKD= ZOKL.
DK = KL; the circle with centre K and radius
KL passes through D.
Similarly this circle passes through E and F.
ZSKD= ZOKL, and SKO is a st. line;
LKD is a st. line
K is the middle point of the hypotenuse of the
rt.Zd A LDK;
the circle with centre K and radius KL passes
through X.
Similarly this circle passes through Y and Z.
the nine points L, M, N, D, E, F, X, Y, Z are
concyclic.
Cor. 1: — The centre of the N.P. circle is the middle
point of the line joining the circumcentre to the ortho
centre.
Cor. 2: — The diameter of the N.P. circle is equal to
the radius of the circumcircle.
simpson s line
Simpson's Line
13
11. If any point is taken on the circumference of
the circumscribed circle of a triangle, the projections
of this point on the three sides of the triangle are
collinear.
Let P be any point on the
circle ABC, X, Y, Z, the pro
jections of P on BC, CA, AB
respectively.
It is required to show that
X, Y, Z are in the same st.
line.
Join ZY, YX, PC, PA.
Z PYC = L PXC, /. P, Y,
X, C, are concyclic, and L
XYC = L XPC.
Z AZP+ Z AYP = 2 rt. Zs,
and Z AYZ = Z APZ.
APCB is a cyclic quadrilateral,
.. Z APC + Z B = 2 rt.
la quadrilateral BZPX,
Zs BZP, BXP arert. Zs,
.. Z ZPX+ z B = 2 rt. Zs.
Hence Z ZPX = Z APC, and as the
common to these Zs,
L APZ = L XPC.
.'. Z AYZ = L XYC, and
L XYC + Z CYZ = L AYZ + CYZ = 2 rt. As \
:. XY and YZ are in the same st. line.
(Historical Notb.— Robert Simpson (16871768) was Professor of Mathematics in
the University of Glasgow.)
/. A, Z, P, Y are concyclic,
Zs.
part APX is
14 SYNTHETIC GEOMETRY
12.— Exercises
(a)
1. P is the orthocentre of A DEF, and the gm EPFG
is completed. Show that DG is a diameter of the circle
circumscribing DEF.
2. ABC is a A ; L, M, N the centres of its escribed
circles. Show that the circle circumscribed about ABC is
the N.P. circle of A LMN.
3. In A ABC, I is the centre of the inscribed circle,
L, M, N the centres of the escribed circles. Prove that the
circumcircle of A ABC bisects IL and LM.
4. O is the orthocentre of A ABC. Prove that As OBC,
ABC have the same N.P. circle.
5. Given the base and vertical Z of a A, show that the
locus of the centre of its N.P. circle is a circle having its
centre at the middle point of the base.
(Note. — Using Cor. 2 of § 10 show that the distance of
K from the fixed point D is constant.)
6. If the projections of a point on the sides of a A are
collinear, the point is on the circumcircle of the A.
7. The three circles which go through two vertices of a
A and its orthocentre are each equal to the circle circum
scribed about the A.
8. The ± from the middle point of a side of a A on
the opposite side of the pedal A bisects that side.
9. Construct a A given a vertex, the circumcircle and
the orthocentre.
(Note. — Describe the circumcircle and produce AO to
cut it in K, where A is the given vertex and O is the ortho
centre. Draw the rt. bisector of OK meeting the circle
in B, C. See § 7, Ex. 6 (c).)
EXERCISES 15
10. DEF is a A and O is its orthocentre. About DOF
a circle is described and EO is produced to meet the
cirumference at P. Show that DF bisects EP.
11. I is the centre of the inscribed circle of A ABC, and
Al, Bl, CI are produced to meet the circumcircle at L, M, N.
Prove that I is the orthocentre of A LMN.
12. I is the centre of the inscribed circle of A ABC, and
the circumcircle of A IBC cuts AB at D. Prove that
AD = AC.
13. In A ABC, the _Ls from A, B to the opposites sides
meet the circumcircle at D, E. Show that arc CD = arc CE.
14. XYZ is the pedal A of A ABC. Prove that A, B, C
are the centres of the escribed circles of A XYZ.
(*)
15. X, Y, Z are the projections of A, B, C on BC, CA,
AB. Prove that
(a) YZ . ZX = AZ . ZB ;
(b) YZ . ZX . XY = AZ . BX . CY.
16. Given the base and vertical L of a A, to find the
loci of the centres of the escribed circles. Let BC be the
base and BDC a segment of a circle containing L BDC =
the given vertical L. Draw the rt. bisector of BC cutting
the circle BDC at D, E. Prove the loci are arcs on the
chord BC and having their centres at D, E.
17. Construct a A having given the base, the vertical Z
and the radius of an escribed circle. (Two cases.)
(Note.— Construct the loci as in Ex. 16. In one case, on
the arc with centre E, find the point l a such that its dis
tance from BC equals the given radius. Draw l x E and
produce to cut the arc BDC at A. ABC is the required A.
16
SYNTHETIC GEOMETRY
In the other case on the arc with centre D find the
point l 2 such that its distance from BC equals the given
radius. Draw l 2 D cutting the arc BDC at A'. A'BC is
the required A.)
18. O is the orthocentre of A ABC, and D, E, F are the
circumcentres of As BOC, COA, AOB. Show that
(a) A DEF = A ABC;
(b) The orthocentre of each of the As ABC, DEF is
the circuincentre of the other ;
(c) The two As have the same N.P. circle.
(Note. — The rt. bisectors of AO, BO, CO form the
sides of A DEF. Let L be the middle point of AO. J_s
from D, E on BC, CA respectively meet at S the circum
centre of A ABC. DS bisects BC at G. Prove L LEO =
L OCA = L GCS and comparing As OLE, GSC, using §8,
show LE = GC. Similarly LF = BG, so that FE = BC, etc.)
19. Find a point such that its projections on the four
sides of a given quadrilateral are collinear.
,'P^
(Note. — ABCD the given quadrilateral; produce the
opposite sides to meet at E, F. Describe circles about
As EBC, FCD meeting again at P. P is the required point.
EXERCISES
17
20. In the A ABC, the _L from A to BC is produced to
cut the circumcircle at P. Prove that the Simpson's Line
of P is  to the tangent to the circumcircle at A.
21. P is any point on the circumcircle of A ABC. The
J_s from P to the sides of the A meet the circle at D, E, F.
Prove that A DEF = A ABC.
22. P is any point on the circumcircle of a A ABC of
which O is the orthocentre and X the projection of A on
BC ; AX produced cuts the circumcircle at D and PD cuts
BC at E. Prove that the Simpson's Line of P bisects PE,
is  OE, and bisects OP.
(Note. — ML cuts PE at F.
Draw PC.
Z FPL = Z FDA = Z PCA =
Z PLF;
/. FL = FP. Then F is the
middle point of the hypotenuse
of the rt.Zd A PLE.
.*. Simpson's Line bisects PE.
V ZFLP= ZXDE; .*. Z FLE
= Z XED. But Z OEX= L XED
(See § 7, Ex. 6 (c).
.'. L OEX = Z FLE, and LM [ OE.
In A POE, LM  OE and bisects PE.: .'. LM bisects PO.)
Give a general statement of the last of the three results
in Ex. 22.
18 synthetic geometry
Areas of Rectangles
13. If from the vertex of a triangle a straight line
is drawn perpendicular to the base, the rectangle
contained by the sides of the triangle is equal to
the rectangle contained by the perpendicular and
the diameter of the circumcircle of the triangle.
AX ± BC and AD is a diame
ter of the circumcircle of A ABC.
To prove that
rect. AB . AC = rect. AX. AD.
Join DC.
V Z AXB = Z ACD,
and Z ABX = Z ADC.
,\ A AXB HI AACD.
AB AX
no. ib. • = .
AD AC
.. rect. AB . AC = rect. AX . AD.
H. If the vertical angle of a triangle is bisected
by a straight line which also cuts the base, the
rectangle contained by the sides of the triangle is
equal to the rectangle contained by the segments
of the base together with
the square on the straight
line which bisects the angle.
ABC is a A and AD the
bisector of L A.
It is required to show b 1
that the rect. AB . AC = rect.
BD . DC + AD 2 .
Circumscribe a circle about
the A ABC. Produce AD to
cut the circumference at E. Join EC.
AREAS OF RECTANGLES
19
111 As BAD, EAC Z BAD = Z EAC, Z ABD = Z AEC,
.*. Z ADB = Z ACE and the As are similar;
i BA EA
heDCe AD = AC'
and .\ BA . AC = AD . EA.
But AD . EA = AD (AD + DE)
= AD 2 + AD . DE
= AD 2 + BD. DC.
.*. rect. BA . AC = rect. BD . DC + AD'.
15. Ptolemy's Theorem: — The rectangle contained
by the diagonals of a quadrilateral inscribed in a
circle is equal to the sum of the rectangles contained
by its opposite sides.
This theorem is a particular case
of that of % 16.
A BCD is a quadrilateral in
scribed in a circle.
To prove that
AC . BD = AB . CD + BC . AD.
Make Z BAE = Z CAD, and
produce AE to cut BD at E.
A ABE U A ACD,
AB _ BE
AC CD'
AB.CD = AC. BE.
A ADE III A ABC,
DE
BC*
FiO. 20.
AD
AC
AD.BC = AC. DE.
AB.CD + AD. BC
= AC. BE + AC . DE
= AC (BE + ED)
= AC . BD.
(Historicai, Note.— Ptolemy, a native of Egypt, flourished in Alexandria in
A.l>.)
SYNTHETIC GEOMETRY
16. The sum of the rectangles contained by the
opposite sides of a quadrilateral is not less than the
rectangle contained by the diagonals.
ABCD is a quadrilateral,
AC, BD its diagonals.
Required to show that
AB . DC + AD . BC is not
less than AC . BD.
Make ^ BAE = z CAD
and Z ADE = L ACB.
Join EB.
As BAC, EAD are similar,
L CAD,
.*. As BAE, CAD are also similar.
From the similar As BAE, CAD
and .*. AB . CD = AC . BE.
AB
BE
AC
CD
From the similar As BAC,
BC ED
AC ~ AD'
id .*. BC
EAD,
AD =
AC . ED.
Consequently AB . CD + BC . AD = AC (BE 4 ED)
but BE + ED is not < BD ;
.*. AB , CD 4 BC . AD is not < AC . BD,
EXERCISES
21
J&^l
17.— Exercises
(a)
1. If the exterior vertical Z A of A ABC be bisected by
a line which cuts BC produced at D, rect. AB . AC = rect.
BD . CD  AD 2 .
(Note. — Draw the circle
ACB. Produce DA to cut
the circumference at E. Draw
EC. ProveABAD ;;ZaEAC,
BA EA _.
and that ,\ ,=: = Tq* lhen
BA . AC = AD . EA = AD
(ED  AD) = AD . ED  AD 2
= BD. CD  AD 2 .)
2. Draw A ABC having a — 81 mm., b = 60 mm., c = 30
mm. Bisect the interior and exterior Zs at A and produce
the bisectors to meet BC and BC produced at D and E.
Measure AD, AE ; and check your results by calculation.
3. If the internal and external bisectors of Z A of A ABC
meet" 3C at L, M respectively, prove
LM 2 = BM . MC  BL. LC.
4. If R is the radius of the circumcircle and A the area
of A ABC, prove that
abc
R =
4 A
If the sides of a A are 39, 42, 45, show that R = 24f.
5. P is any point on the circumcircle of an equilateral
A ABC. Show that, of the three distances PA,
one is the sum of the other two.
PB, PC,
(»)
6. From any point P on a circle ±s are drawn to the
four sides and to the diagonals of an inscribed quadrilateral.
Prove that the rect. contained by the ±s on either pair of
82
SYNTHETIC GEOMETRY
opposite sides is equal to the rect. contained by the J_s on
the diagonals.
7. "With given base and vertical Z construct a A having
the rect. contained by its sides equal to the square on a
given st. line.
8., A, B, C, D are given points on a circle. Find a point
P on the circle such that PA . PC = PB . PD.
9. AB is the chord of contact of tangents drawn from a
point P to a circle. PCD cuts the circle at C, D. Prove
that AB . CD = 2 AC . BD.
10. I is the centre of the inscribed circle of A ABC. Al
produced meets the circuracircle at K. Prove Al. IK = 2Rr.
(Note. — Draw the diameter KE of
circle ABC and the radius IN of the
inscribed circle. Draw BE, BK, Bl.
Show that A BEK III A NAI, and
rt. * . Al KE 
that .. W = Kg.
or, Al . KB = 2Rr.
Show that KB = Kl, and that .«.
Al . IK = 2Rr.)
Hence, using Ex. 6, Page 256,
O.H.S. Geometry, show that, if S be the circumcentre of
A ABC, SI 2 = R 2  2Rr.
11. I x is the centre of the escribed circle opposite to A
in A ABC. Ah cuts the circumcircle ABC at K. Prove
Al 2 . I X K = 2Rr x .
Hence show that, if S be the circumcentre of A ABC,
Sli 3 = R 2 + 2R/v
RADICAL AXIS
23
Radical Axis
18. The locus of the points from which tangents
drawn to two circles are equal to each other is called
the radical axis of the two circles.
19. If two circles cut each other, their common
chord produced is the radical axis.
[Proof left for the pupil]
20. The locus of a point P
such that the difference of the
squares of its distances from
two fixed points A, B is
constant is a st. line perpen
dicular to AB. FlQ 25
From P draw PM j_ AB. Let AB = a, AM
PA 2  PB 2 = k, where a and k are constants.
AM 2 4 MP 2 = PA 2
MB 2 +MP 2 = PB 2
.*. AM 2  MB 2 = PA 2  PB 2 = h.
or x 2  (a  x) 2 = k.
a 2 +k
and
.
and x =
2a
24
SYNTHETIC GEOMETRY
Hence AM is constant and M is a fixed point.
.*. the locus of P is a st. line 1 AB drawn through
the fixed point M.
21. A, B are the centres of two circles of radii
R, r respectively.
To prove that the radical axis of the circles is a
st. line _L AB and cutting it at a point M such that
AM 2  MB 2 = R 2  r\
Let P be any point on the radical axis, and draw
PM ± AB.
Draw the tangents PC, PD to the circles, and join
PA, PB, AC, BD.
PA 2 = PC 2 + R 2 ,
PB 2 = PD 2 + r 2 ,
and, since P is on the radical axis, PC = PD;
.. PA 2  PB 2 = R 2  r 2 , a constant.
.•., by § 20, the locus of P is a st. line _L AB.
Also PA 2 = AM 2 +PM 2 ;
PB 2 = MB 2 +PM 2 l
.♦. PA 2  PB 2 = AM 2  MB 2 ;
and .'. the radical axis cuts AB at the fixed point M,
such that AM 2  MB 2 = R 2  r 2 .
EADICAL AXIS
25
22. To draw the radical axis of two noninter
secting circles.
Let A, B be the centres of the circles.
Describe a circle with centre O cutting the given
circles at C, D and E, F. Draw CD, EF and produce
them to meet at P. Draw PM j_ AB. Draw tangents
PG, PH to the circles.
Show that PM is the required radical axis.
SYNTHETIC GEOMETRY
Second Method
Let A, B bo the centres of the two circles.
Join AB. Through B draw BC _L AB cutting the
circle with centre B at C, and cut off BD equal to
the radius of the other circle.
With centre A and radius AD describe an arc, and
with centre B and radius AC describe another arc
cutting the first at E. Draw EM J_ AB.
BM 2 = BE 2  EM 2 = AB 2 + BC 2  EM 2 .
MA 2 = AE 2  EM 2 = AB 2 + BD 2  EM 2 .
.*. BM 2  MA 2 = BC 2  BD 2 .
•'• > by § 21, EM is the radical axis.
EXERCISES 27
23.— Exercises
1. Draw two circles, radii 1 inch and 2 inches, with
their centres 4 inches apart. Find a point whose tangents
to the two circles are each 1 inches in length.
2. The radical axis of two circles bisects their common
tangents.
3. Find the radical axis of two circles which touch each
other, internally or externally.
i. Prove that the radical axes of any three circles taken
two and two together meet in a point.
Note. — This point is called the radical centre of the
three circles.
5. O is a fixed point outside a given circle. P is any
point such that the tangent from P to the given circle =
PO. Show that the locus of P is a st. line J_ to the line
joining O to the centre of the circle.
<& C is a point on the circumference of a circle with
centre A. Join AC and draw CB J_ CA. "With centre B
and radius BC describe a circle.
(a) The tangents to the circles at a common point
are L to each other.
(b) The square on the line joining the centres of the
circles equals the sum of the squares on their radii.
Definition. — Circles which cut each other so that the Js
tangents at a common point are at right angles to each
other are said to be orthogonal.
7. If O be the orthocentre of A ABC, the circles described
on AB and CO as diameters are orthogonal.
8. Tf circles are described on the three sides of a A as
diameters, their radical centre is the orthocentre of the A.
28
SYNTHETIC GEOMETRY
9. Through two given points A and B draw any number
of circles. What is the locus of their centres? Show that
any two of this system of circles have the same st. line for
radical axis.
Definition — A system of circles that have the same
radical axis are said to be coaxial.
10. To draw a system of circles coaxial with two given
nonintersecting circles.
Let A, B be the centres of the given circles. Draw their
radical axis PO cutting AB at O. From O draw a tangent
OE to either circle. With centre O and radius OE describe
a circle cutting AB at C, D. On the circle CDE take any
point F and at F draw a tangent to the circle CED cutting
AB at G. With centre G and radius GF describe a circle.
Prove that this circle is coaxial with the given circles.
By taking different positions of F on the circle CED any
number of circles may be drawn coaxial with the given
oircles.
Definition. — No circle of the coaxial system has its centre
between C and D, and consequently these points are called
the limiting points of the system.
EXERCISES 29
11. In Fig. 29, show that :—
(a) The circle CED cuts each circle of the coaxial
system orthogonally;
(b) Any circle with centre in PO and passing through
C, D cuts any circle of the coaxial system ortho
gonally.
(*)
12. If from any point P tangents be drawn to two circles,
the difference between their squares equals twice the rect
angle contained by the _L from P on the radical axis of
the two circles and the distance between their centres.
13. The difference of the squares of the tangents drawn
from a point to two fixed circles is constant. Show that
the locus of the point is a st. line ± to the line of centres
of the circles.
14. The tangent drawn from a limiting point to any
circle of a coaxial system is bisected by the radical axis.
15. Show that the locus of the centre of a circle, the
tangents to which from two given points are respectively
equal to two given st. lines, is the radical axis of the circles
having the given points as centres and radii respectively
equal to the two given st. lines.
16. With a given radius describe a circle to cut two
given circles orthogonally.
17. XYZ is the pedal A of A ABC; YZ, BC meet in L;
ZX, CA meet in M ; XY, AB meet in N. Show that L, M, N,
are on the radical axis of the circumscribed and N.P.
circles of A ABC.
(Note. — As MAZ, MXC are easily shown to be similar.)
18. Describe a circle to cut three given circles orthogonally.
(Note.— Use Ex. 4 and Ex. 11 (&).)
CHAPTER II
Medial Section
,53". When a straight line is divided into two parts
such that the square on one part is equal to the
rectangle contained by the
"**"v**< v given straight line and the
\ \ other part, the straight line
is said to be divided in
j '• \ medial section.
S F ..•'" (25^ To divide a given
„..**" straight line internally in
i ..'' medial section.
D;'" Let AB be the given st.
: line.
Draw AC j_ AB and = AB.
£■ Bisect AC at D. With centre
Fl ° 30  D and radius DB describe an
arc cutting CA produced at E. With centre A and
radius AE describe an arc cutting AB at F.
Then AB is divided in medial section at F.
DA 2 +AB 2 = DB 2
= DP
= DA 2 + 2 DA. AE + AE 2
= DA 2 + AB. AF + AF 2 ;
V 2 DA = AC = AB and AE = AF.
.*. AF 2 = AB 2  AB . AF.
= AB (AB  AF)
= AB . BF.
30
MEDIAL SECTION
31
26. To divide a given straight line externally in
medial section.
Let AB be the given st. line.
A
D
C
E ,.
^'B\
Draw AC _L AB and = AB. Bisect AC at D. With
centre D and radius DB describe an arc cutting AC
produced through C at E. With centre A and radius
AE describe an arc cutting BA produced through
A at F.
Then AB is divided externally at F such that
AF 2 = AB . BF.
DA 2 + AB 2 = DB 2
= DE 2 = (AE  AD) 2
= AE 2  2 AE. AD + AD 2
= AF 2  AF. AB + AD 2
AF 2 = AB 2 + AF. AB
= AB (AB + AF)
= AB . BF.
32
SYNTHETIC GEOMETRY
27.— Exercises
1. If a st. line AB be divided at F so that AP= AB.BF,
show that AB : AF = AF : BF.
Give a general statement of this result.
2. AB is divided internally at F such that AF 3 = AB . BF.
Show that AF > FB.
3. AB is divided in medial section at
F. On AB, AF squares ABCD, AFEG
are described as in Fig. 32. EF is pro
duced to meet DC in H. Show that
the rectangle DE = the square DB.
4. In Fig. 32 join GB, DF, pro
duce DF to meet GB, and show that
DF _[_ GB.
5. A given st. line AB is to be
divided in medial section. Let F be
Fl0 3 2 the point of section, a the length of
AB, x the length of AF.
Then, by the definition of medial section, x 2 = a (a — x)
or, x f ax  a 2 = 0.
c , . ,, . n . .  a ± a i/5
feolving this quadratic equation, a; = •
G
E
F
A
D
H
Show that the construction in § 25 is sugg
a \ a V5
a  a /5
ted by the root
and the construction in § 26 by the root
6. Divide a st. line 4 inches in length in medial section.
Measure the length of each part, and test the results by
calculation.
7. The difference of the squares on the parts of a st. line
divided in medial section equals the rectangle contained by
the parts.
MEDIAL SECTION 33
'8. If AB be divided at C so that AC 2 = AB . BO, show
that AB 2 + BC 2 = 3 AC 2 .
'9. If the sides of a rt. Zd A are in continued proportion,
the J from the rt. Z divides the hypotenuse in medial
section.
10. Describe a rt.Zd A "whose sides are in geometrical
progression.
28. To describe an isosceles triangle having each
of the angles at the base double the vertical angle.
Draw a st. line AB and divide it at H so that
AH' = AB.BH. (§25.)
Describe arcs with centres A, B and radii AB, AH
respectively, and let them cut at C.^^
Join AC, BC. /A
ABC is the required A.
Join HC.
Z B is common to the As ABC, CBH,
and since
AB AH . AB BC
AH BH' •'• BC BH'
B C
FlQ. 33.
.*. these As are similar and z BCH =
= Z A.
. . AC A B AH
A g ain BC = AH = HB'
CH bisects Z ACB.
Z ACB = twice Z BCH = twice Z A;
and also Z ABC = twice Z A.
34
SYNTHETIC GEOMETRY
L'9.— Exercises
(«)
1. Express the Zs of the A ABC (Fig. 33) ia degrees.
2. Construct Zs of 36°. 18°, 9°, 6°, 3°. $1°
* 3. Show that AHC (Fig. 33) is an isosceles A having the
vertical Z three times each of the base Zs.
i. Show that each Z of a regular pentagon is 108° ; and
that, if a regular pentagon is inscribed in a circle, each
side subtends at the centre an Z of 72°.
* 5. In a given circle CAB draw
any radius OA. Divide AO at H
so that OH 2 = AO . AH. Place the
chord AB = HO.
Join BH and produce BH to cut
the circumference at C. Join AC.
Show that AB is a side of a
regular decagon inscribed in the
circle ; and that AC is a side of
a regular pentagon inscribed in
the circle.
J6J In a given circle inscribe a regular pentagon. At the
angular points of the pentagon draw tangents meeting at
A, B, C, D, E. Show that ABODE is a regular pentagon
circumscribed about the circle.
/T. Show that each diagonal of a regular pentagon is 
to one of its sides.
8. Draw a regular pentagon on a given st. line.
9. ABCDE is a regular pentagon. Show that AD, BD
trisect Z CDE.
EXERCISES 35
10. ABCDE is a regular pentagon. Show that AC, BD,
divide each other in medial section.
11. Construct a regular 5pointed star. What is the
measure of the Z at each vertex?
0)
12. Show that the side of a regular decagon inscribed in
f
a circle of radius r is — (]/ 5  1).
13. Show that the side of a regular pentagon inscribed in
a circle of radius r is — »/ 10  2y 5.
14. The square on a side of a regular pentagon inscribed
in a circle equals the sum of the squares on a side of the
regular inscribed decagon and on the radius of the circle.
15. In a circle of radius 2 inches inscribe a regular decagon
by the method of Ex. 5. Measure a side of the decagon and
check your result by calculation.
16. In a circle of radius 3 inches inscribe a regular pentagon
by the method of Ex. 5. Measure a side of the pentagon and
check your result by calculation.
17. In a given circle draw two radii OA. OB at rt. Zs to
each other. Bisect OB at C. ^"""~^"**"»>^
Join AC, and cut off CD = CO. yf ^v
Show that AD is equal to a / \
side of a regular decagon in / \
scribed in the circle. [~""" r "P
The regular inscribed pentagon \ ;\ / J
may be drawn by joining alter \ f / /
nate points obtained by placing N. ;/ yf
successive chords each equal to ^
AD. Fio. 35.
18. On a st. line 2 inches in length describe a regular
pentagon. Measure a diagonal of the pentagon and check
your result by calculation.
db SYNTHETIC GEOMETRY
19. If the circumference of a circle be divided into n
equal arcs,
(a) The points of division are the vertices of a regular
polygon of n sides inscribed in the circle;
(6) If tangents be drawn to the circle at these points,
these tangents are the sides of a regular polygon
of n sides circumscribed about the circle.
20. Show that the difference between the squares on a
diagonal and on a side of a regular pentagon is equal to
the rectangle contained by them.
tn/3f>*n/°C
MISCELLANEOUS THEOREMS 37
Miscellaneous Theorems
30. ABC is a triangle and P is a point in BC
such that !!£ = —. It is required to show that
PC ni
mAB 2 + nAC* = (m + n) AP 2 4 mBP 2 + nCP\
Draw AX jl BC.
From A ABP, /i\V
AB 2 = AP 2 + BP 2  2 BP.PX. /] VV
From A APC, / j \ \
AC 2 = AP 2 +CP 2 +2 CP, PX. B X p Q
Fig. 36.
Multiplying both sides of
the first of these equations by m, both sides of the
second by n, adding the results and using the condition
??i BP = tiPC, we obtain
mAB 2 + nAC 1 = (m + n) AP 2 + mBP 2 + ?iCP 2 .
What does the result in § 30 become when m = n ?
In a A ABC, a = 77 mm, 6 = 90 mm and c = 123
mm. Find the distances from C to the points of
trisection of AB.
(31. In a rightangled triangle a rectilineal figure
described on the hypotenuse equals the sum of the
similar and similarly described figures on the other
two sides.
ABC is a A rt.Zd at C and having the similar and
similarly described figures X, Y, Z on the sides.
It is required to show that X + Y = Z.
38
SYNTHETIC GEOMETRY
Similar figures are to each other as the squares on
corresponding sides.
Y
•• Z
AC 2
~ AB''
and =
BC 2
~ AB 2 '
X + Y
Z
AC 2 +BC 2
AB 2
But
AC 2 + BC 2
= AB 2 .
.'. X + Y
= Z.
Prove this theorem by drawing a J_ from C to AB
and using the theorem : — If three st. lines are in
continued proportion, as the first is to the third so is
any polygon on the first to the similar and similarly
described polygon on the second.
SIMILAR AND SIMILARLY SITUATED POLYGONS 39
Similar and Similarly Situated Polygons
f Similar polygons are said to be similarly situ
when their corresponding sides are parallel and
drawn in the same direction from the corresponding
vertices.
33. If two similar triangles have their corresponding
sides parallel, the st. lines joining corresponding vertices
are concurrent.
Let ABC, DEF be two similar As having the sides
BC, CA, AB respectively  to the corresponding sides
EF, FD, DE. ~ oc.or
c r
Prove AD, BE, CF concurrent.
('34.; When two similar polygons are so situated that
their corresponding sides are parallel but drawn in
opposite directions from the corresponding vertices,
they are said to be oppositely situated.
In Fig. 39, the similar As ABC, DEF are oppositely
situated.
40
SYNTHETIC GEOMETRY
35. If two similar polygons have the sides of one
respectively parallel to the corresponding  sides of
the other, the straight lines joining corresponding
vertices are concurrent.
Let ABCDE, abcde be two similar polygons, similarly
situated in Fig. 40, oppositely situated in Fig. 41.
Join Aa, B6 and let the joining lines meet at O.
Join CO cutting he at x.
From similar As,
ah
AB
Oh
OB
hx
= BC
EXERCISES 41
But, by hypothesis,
ab be
AB = BC*
bx = be, and
OC passes through c.
Similarly it may be shown that the st. lines joining
the remaining pairs of corresponding vertices pass
through O.
36.— Exercises
(a)
X. Inscribe a square in a given A. Show that there are
three solutions.
(Note. — In A ABC on BC externally describe the square
BDEC. Draw AD, AE cutting BC at F, G respectively.
Draw FH, GK _L BC meeting BA, CA at H, K respectively.
Draw HK. Prove that HFGK is a square.)
2. In a given A inscribe a rectangle similar to a given
rectangle. Show that there are six solutions.
3. In a given semicircle inscribe a square.
"i Ina given semicircle inscribe a rectangle having its
sides in a given ratio.
"^ In a given A inscribe a A having its sides  to three
given st. lines.
(Note. — From any point D in the side BC of the
given A ABC draw DE  to one of the given st. lines and
meeting AC at E. Draw DF, EF respectively  to the other
given lines. Draw CF cutting AB at G. Draw GH  FD
and GK  FE, meeting BC, AC respectively at H, K. Draw
HK. Show that GHK is the required A.)
(*)
6. The base of a square lies on one given st. line and one
of its upper vertices lies on another given st. line. Show that
the locus of the other upper vertex is a fixed st. line passing
through the point of intersection of the two given lines.
42 SYNTHETIC GEOMETRY
7. In Figures 40 or 41 P is any point in AB, Q is any
point in CD and p, q are the corresponding points in ab,
cd respectively. Prove that PQ  pq.
8. A ABC HI A «& c » with their corresponding parts in the
same circular order, but their corresponding sides are not .
BC, be meet at P. The circles BPb, CPc meet again at O.
Prove that the A abc may be rotated about O to a position
where it is similarly situated to A ABC.
(Note.— Prove L BOb = Z COc  L AOa.)
Fig. 42.
9. Show that 4R = r x \r 2 \r 3  r, when R is the radius
of the circumcircle, r of the inscribed circle, and r v r.,, r 3 the
radii of the escribed circles.
(Note. — In A ABC let I be the centre of the inscribed
circle, lj that of the escribed circle opposite A. Draw the
diameter GDH of the circumcircle bisecting BC at D, and
meeting the circle above BC at G below at H. Draw IM,
IjN _L BC and produce GH to cut Mlj at K. Prove
2 GD = r 2 \r 3 and 2 DH = r t  r.)
10. If I, m, n are the _Ls from the circumcentre on the
sides of a A, show that
l\m\n = R + r.
(Note. — Use the diagram and result of Ex. 9.)
CHAPTER III
Harmonic Ranges and Pencils
37. A set of collinear points is called a range.
Q8< A set of concurrent straight lines is called a pencil.
The lines are called the rays of the pencil ; and
their common point is called the vertex of the pencil.
39. When three magnitudes are such that the first
has the same ratio to the third that the difference
between the first and second has to the difference
between the second and third, the differences being
taken in the same order, the magnitudes are said to
be in harmonic proportion. (H. P.)
Thus, if a, b and c represent three numbers such
that a : c = b  a : c  b, a, b and c are in H. P.
'40. If in a range of four points A, c. b, d the st.
line AB is divided internally at C and externally at
D in the same ratio, the distances from either end
of the range to the other three points are in H. P.
/
A
c/
B
D
J
/
jt"
'
^~"
/
,'(
*'"
Draw is AE, FBG to AB, making FB = BG. Draw
EF, EG cutting AB at C, D.
Then AC
CB
AE
BF
AE
BG
AD
DB
44
SYNTHETIC GEOMETRY
v
.D
(a) Then since AC : CB = AD : DB.
by alternation, AC : AD = CB : DB.
/. AC : AD = AB  AC : AD  AB.
.*. by the definition of § 39, AC, AB, AD are in H. P.
(b) By inversion, DB : AD = BC : AC.
;. DB : DA = DC  DB : DA  DC.
•'• , by § 39, DB, DC, DA are in H. P.
State and prove a converse to this theorem.
Gfcli When a range of four points A, C, B, D is such
that AC : CB = AD : DB, it is called a harmonic range.
If any point P be joined to the four points of a har
monic range, the joining lines form a harmonic pencil.
Tl If, in the harmonic pencil P (A, c, B, D), a
straight line through B parallel to PA cut PC, pd
at E, F respectively, BE = BF.
^P
Fig. 44.
V AACP]ABCE,
.'. AP : EB = AC : CB.
V AADPIHABDF,
.'. AP : BF = AD : DB.
But, by hypothesis, AC :CB = AD : DB.
/. AP : EB = AP : BF.
EB = BF.
HARMONIC RANGES AND PENCILS 45
(43) By a proof similar to that of § 42, the following
converse to the theorem of that article may be shown
to be true.
If, in the pencil p(a, c, b, D), a straight line
through B parallel to PA cut PC, PD at E, F re
spectively such that be = bf, then P(a, c, b, d) is a
harmonic pencil.
44. Any transversal is cut harmonically by the
rays of a harmonic pencil.
A transversal cuts the rays PA, PC, PB, PD of the
harmonic pencil P(A, C, B, D) at K, M, N, L respectively.
Fia. 45.
It is required to show that K, M, N, L is a harmonic
range.
Through B, N respectively draw EF, GH  PA.
V P(A, C, B, D) is a harmonic pencil, and EBF  AP
.. , by § 42, EB = BF.
™ • '1 a GN EB 1 NP BP
From similar As, ^ = — and — = —,
.. , by multiplication, GN : NH = EB : BF,
.. GN = NH; and /., by § 43, K, M, N, L is a
harmonic range.
46 SYNTHETIC GEOMETRY
middle
[f A,
poii
or,
C, B ;
it of
, D is a harmonic range and O is the
AB,
OB 2 = OC . OD.
?cc3
and
A
O C B
Fio. 46.
AC AD
CB~DB'
AC + CB AD+DB
AC  CB ~ AD  DB'
20B 20 D
20c ^OB'
OB 2 = OC . OD.
D
£<S2> ^ £
/State and prove a converse to this theorem.
46. If A, C, B, D is a harmonic range, A and B are
said to be harmonic conjugates with respect to C
and D ; and C and D are said to be harmonic con
jugates with respect to A and B.
47.— Exercises
(a)
1. Show how to find the fourth ray of a harmonic pencil
when three rays are given.
2. Prove the theorem of § 44 when the transversal cuts
the rays produced through the vertex.
3. In the A ABC the bisectors of the interior and
exterior ^s at A cut BC and BC produced at D, E
respectively. Show that B, D, C, E is a harmonic range.
s£. A, C, B, D is a harmonic range and P is any point
on the circle described on AB as diameter. Show that
PA, PB respectively bisect the exterior and interior vertical
Zs of A CPD.
(Note. — Draw EBF II AP cutting PC, PD at E, F re
spectively.)
EXERCISES 47
5. Using Ex. 4, give geometrical proofs of the converse
theorems of § 45.
/6T Two circles cut orthogonally. A st. line through the
centre of either cuts the circles at A, C, B, D. Show that
A, C, B, D is a harmonic range.
/K. A, C, B, D is a harmonic range. Show that the
circles described on AB, CD as diameters cut each other
orthogonally.
(b)
8. A, C, B, D is a harmonic range ; O is the middle
point of AB and R is the middle point of CD. Show that: —
(a) AB 3 4 CD 2 = 4 OR';
(b) CA. CB = CD . CO;
(c) If P is any point on the range and lengths in
opposite directions from P are different in sign,
PA . PB + PC . PD = 2 PO . PR.
Jf. The inscribed circle of A ABC touches BC, CA, AB
at D, E, F respectively, and DF meets CA produced at P.
Show that C, E, A, P is a harmonic range.
(Note. — Use Menelaus' Theorem.)
10. The diameter AB of a circle is ± to a chord CD.
P is any point on the circumference. PC, PD cut AB, or
AB produced, at E, F. Show that A, E, B, F is a harmonic
range.
11. Through E, the middle point of the side AC of the
A ABC, a transversal is drawn to cut AB at F, BC
produced at D, and a line through B  CA at G. Show
that g, F, E, D is a harmonic range.
(Note.— Use § 43.)
48 SYNTHETIC GEOMETRY
12. A common tangent of two given circles is divided
harmonically by any circle which is coaxial with the given
circles.
(Note. — Use the converse of § 45.)
13. In a circle AC, BD are two diameters at rt. Z.s to
each other, and P is any point on the quadrant AD. Show
that PA, PB, PC, PD constitute a harmonic pencil.
14. In the harmonic pencil O (A, C, B, D), Z AOC =
Z COB = Z BOD. Show that each of these Zs = 45°.
15. A square is inscribed in a circle. Show that the
pencil formed by joining any point on the circumference to
the four vertices of the square is harmonic.
16. P is a fixed point and OX, OY two fixed st. lines.
Show that the locus of the harmonic conjugate of P with
respect to the points where any st. line drawn from P cuts
OX, OY is a st. line passing through O.
THE COMPLETE QUADRILATERAL
49
The Complete Quadrilateral
@. The figure formed by
four straight lines which meet
in pairs in six points is called
a complete quadrilateral.
The figure ABCDEF is a
complete quadrilateral, of
which AC, BD and EF are
the three diagonals.
(49, In a complete quadrilateral each diagonal is
divided harmonically by the two other diagonals,
and the angular points through which it passes.
ABCDEF is a complete quadrilateral having the
diagonal AC cut by DB at P and by EF at Q.
It is required to show that A, P, C, Q is a harmonic
range.
In A ACF, AB, CD, FQ drawn from the vertices and
meeting the opposite sides at B, D, Q are concurrent at
E and .*. , by Ceva's Theorem,
FD. AQ. CB = DA . QC. BF.
*■?*.%'
50 SYNTHETIC GEOMETRY
The transversal DPB cuts the sides of the
A ACF at
D, P, B and /. , by Menelaus' Theorem,
FD . AP . CB DA . PC . BF.
V
Hence, by division,
AQ _ Q^,
AP PC f
\
AP AQ
or ' PC ~ QC
and .. A, P, C, Q, is a harmonic range.
From the above result it is seen that F (A, P, C, Q)
is a harmonic pencil, and consequently, by § 44,
D, P, B, R is a harmonic range.
Show, in the same manner that F, Q, E, R is a
harmonic range.
Poles and Polars
50. If through a fixed point a line be drawn to cut
a given circle and at the points of intersection tangents
be drawn, the locus of the intersection of the tangents
is called the polar of the fixed point; and the fixed
point is called the pole of the locus.
POLES AND POLARS
51
51. If c is the centre of a given circle and d is a
fixed point, the polar of D with respect to the
circle is a straight line which is perpendicular to
CD and cuts it at a point E such that CE . CD equals
the square on the radius.
Fig. 49. Fig. 60.
Through D draw any st. line cutting the circle at
A and B. At A, B draw tangents to the circle inter
secting at P.
P is a point on the polar of D.
Join CD and from P draw PE J_ CD.
Join CP cutting AB at F. Join CB.
V Zs DEP, DFP are rt. Zs,
.*. D, E, F, P are concyclic.
.*. CE . CD = CF . CP.  P&C^'*
But CF . CP = CB'. 7JP)
.'. CE^.CD = CB'.= r i
Then, since CD and CB^are constants, CE must also
be constant.
.. the polar of D must be the st. line 1CD through
the fixed point E such that CE . CD = CB 2 .
52 SYNTHETIC GEOMETRY
J>2. If a point P lies on the polar of a point Q
with respect to a circle, then Q lies on the polar
of P.
P is any point on PM
the polar of Q.
To show that Q lies
on the polar of P.
Join CP and draw
QN X CP.
V Zs at M and N
are rt. Zs.
.*. Q, N,
concyclic.
.. CP . CN = CM . CQ but, by § 51, CM
square on the radius.
.*. CP . CN = the square on the radius, and QN is
the polar of P.
53. Two points, as P and Q in Fig. 51, such that the
polar of each passes through the other, are called
conjugate points with respect to the circle, and their
polars are called conjugate lines.
A A such that each side is the polar of the oppo
site vertex is said to be selfconjugate.
P, M are
CQ = the
EXERCISES 53
54.— Exercises
(«)
Y? P is a point at a distance of 4 cm. from the centre
of a circle of radius 6 cm. Construct the polar of P.
X P is a point at a distance of 7 cm. from the centre
of a circle of radius 5 cm. Construct the polar of P.
3. Draw a st. line at a distance of 7 cm. from the centre
of a circle of radius 4 cm. Construct the pole of the line.
/£. When the point P is within the given circle, the
polar of P falls without the circle ; and when P is without
the circle, the polar of P cuts the circle.
5. The polar of a point on the circumference is the
tangent at that point.
9f. P is a point without a given circle and the polar of
P cuts the circle at A. Show that PA is a tangent to the
circle.
Give a general statement of this theorem.
^ If any number of points are collinear, their polars
with respect to any circle are concurrent.
^ Any number of lines pass through a given point; find
the locus of their poles with respect to a given circle.
9l If the pole of a st. line AB with respect to a circle
is on a st. line CD, the pole of CD is on AB.
"KL The st. line joining any two points is the polar with
respect to a given circle of the intersection of the polars
of the two points.
J }A. The intersection of any two st. lines is the pole with
respect to a given circle of the line joining the poles of
the two st. lines.
yi. Show how to draw any number of self con jugate As
with respect to a given circle.
54 SYNTHETIC GEOMETRY
<*)
1/5. If a st. line PAB cut a circle at A, B and cut the
polar of P at C, and if D be the middle point of AB,
PA . PB = PC . PD.
y(. Two circles ABC, ABD cut orthogonally. Show that
the polar of D, any point on the circle ABD, with respect
to the circle ABC passes through E, the point diametrically
opposite to D.
PA, The polar of any point A with respect to a given
circle with centre O cuts OA at B. Show that any circle
through A and B cuts the given circle orthogonally.
r<i, A is a given point and B any point on the polar of
A with respect to a given circle. Show that the circle
described on AB as diameter cuts the given circle ortho
gonally.
17. ABC is a A inscribed in a circle, and a  to AC
through the pole of AB with respect to the circle meets
BC at D. Show that AD = CD.
55. Any straight line which passes through a
fixed point is cut harmonically by the point, any
circle, and the polar of the point with respect to
the circle.
P is the fixed point, O the centre of the circle,
PACB any line through P cutting the circle at A, B
and the polar EC of P with respect to the circle at C.
POLES AND POLARS
55
It is required to show that B, C, A, P is a harmonic
range.
Join BO, OA, BE, EA and produce BE to F.
By § 51, OP . OE = OB 2 , and .'. OP:OB = OB:OE;
also the Z BOE is common to the A POB, BOE;
.'. these A are similar,
and consequently Z OEB  z OBP = Z OAB.
.'. the points B, O, E, A are concyclic.
.*. the Z AEP = Z OBA = Z OEB = Z FEP, and EP
bisects the exterior vertical Z of A EBA.
But Z PEC is a rt. Z and .'. Z BEC = Z CEA.
BC BE BP
" CA ~ EA ~ PA'
and B, C, A, P is a harmonic range.
X If the point P is within the circle, C is without the
circle, and by § 52, the polar of C passes through p.
.'. by the above proof B, P, A, C is a harmonic
range.
k Prove this theorem when the line PAB passes through
the centre of the circle.
56 SYNTHETIC GEOMETRY
56. ABCD is a quadrilateral inscribed in a circle.
AB, DC are produced to meet at E ; BC, AD to meet
at F, forming the complete quadrilateral ABCDEF.
AC cuts BD at G, FG cuts AB at L.
From the complete quadrilateral FDGCAB, A, L, B, E
and D, K, C, E are harmonic ranges, d (§49.)
.*. L and K are points on the polar of E ; (§ 55.)
that is, GF is the polar of E.
Similarly, GE is the polar of F.
Hence FE is the polar of G ; and the A EFG is self
conjugate with respect to the circle ABC.
Cor.: — If from any point E two st. lines EBA, ECD are
drawn to cut a circle at the points B, A, C, D, then
the intersection of BD and CA is on the polar of E, and
so also is the intersection of BC and AD.
EXERCISES 57
57.— Exercises
(«)
X. Using a ruler only, find the polar of a given point with
respect to a given circle.
/S. Using a ruler only, draw the tangents from a given
external point to a given circle.
3. Using a ruler only, find the pole of a given st. line with
respect to a given circle.
4. A and B are two points such that the polar of either
with respect to a circle, with centre O, passes through the
other. Prove that the pole of A B is the orthocentre of the
A AOB.
f. Tangents AB, AC are drawn to a circle. The tangent
at any point P cuts BC, CA, AB at X, Y, Z respectively.
Show that X, Z, P, Y is a harmonic range.
6. If, in figure 53, ABC is a fixed A and D is a variable
point on the circle, prove that each side of the A EFG
passes through a fixed point.
(Note.— Use Ex. 9, § 54.)
Jt: C is the middle point of a chord AB of a circle, and
D, E are two points on the circumference such that CA
bisects the L DCE. Prove that the tangents at D and E
intersect on AB.
(Note. — At C draw the ± to AB and produce it to
meet DE.)
)£ D is the middle point of the hypotenuse BC of the
rt.Zd A ABC. A circle is described to touch AD at A.
Prove that the polar of either of the points B, C with
respect to the circle passes through the other point.
jtf. A, B, C, D are successive points on a st. line. Find
points X, Y that are conjugate to each other both with
respect to A, B and with respect to C, D.
58 SYNTHETIC GEOMETRY
(Note. — Draw a circle through A, B and another through
C, D, intersecting each other at P, Q. Produce PQ to
cut the given line at O and from O draw a tangent OT to
either of the circles. With centre O and radius OT describe
a circle cutting the given line at X, Y.)
10. AD, BE, CF are concurrent st. lines drawn from the
vertices of A ABC and cutting the opposite sides in D, E, F.
EF meets BC at X. Show that X is the harmonic conjugate
of D with respect to B and C.
11. A transversal cuts the sides BC, CA, AB of the
A ABC in D, E, F. The st. line joining A to the inter
section of BE and CF meets BC at H. Show that D and
H are harmonic conjugates with respect to B and C.
12. P, Q, are two conjugate points with respect to a
circle. Show that the circle on PQ as diameter cuts the
given circle orthogonally.
13. The Z C in A ABC is obtuse and O is the ortho
centre. A circle described on OA as diameter cuts BC at
D. Show that the A ABC is selfconjugate with respect to
the circle with centre O and radius OD.
(*)
14. If a quadrilateral be circumscribed about a circle, the
st. lines joining the points of contact of opposite sides are
concurrent with the two diagonals of the quadrilateral.
(Note. — Produce the opposite chords of contact to meet,
and use § 56 — Cor.)
15. If PM, QN be respectively drawn _L to the polars of
Q, P with respect to a circle whose centre is O, PM : QN =
OP : OQ. Salmon's Theorem.
(Note. — Draw OK ± PM, OL _L QN ; and use similar As
QPK, OQL.)
EXERCISES 59
16. D, E, F are points on the sides BC, CA, AB of the
A ABC, such that AD, BE, CF are concurrent. Prove that
the harmonic conjugates of D with respect to B and C, of
E with respect to C and A, and of F with respect to A and
B are collinear.
17. In A ABC, X is the projection of A on BC, and BC
is produced to cut the radical axis of the circumcircle and
the N.P. circle at P. Show that B, X, C, P is harmonic.
18. The opposite sides of the quadrilateral A BCD are
produced to meet at E, F. The diagonals of the complete
quadrilateral form the A LMN. Show that the circle
described on any one of the diagonals as diameter cuts the
circumcircle of A LMN orthogonally.
60
SYNTHETIC GEOMETRY
Maxima and Minima
(58,; If a magnitude, such as the length of a st. line,
an angle, or an area, varies continuously, subject to
given conditions, it is said to be a maximum when it
has its greatest possible value; and a minimum when
it has its least possible value.
(59) Let the distance, PA from a
fixed point P within a circle to the
circumference be the magnitude in
question.
Join A to the centre O and produce
PO to meet the circumference at
B and C. Afl + o# ; ^oy0"B
o»  OP •. pC.
PA < PO + OA but > OA  OP ;
.'. PA < PB but > PC.
As PA rotates about P its length varies continuously.
When it comes to the position PB it is greater than
in the positions close to PB on either side, and has its
maximum value.
Again, at PC it is less than in the positions close to
PC on either side, and has its minimum value.
Draw the diagram and illustrate in the same manner
the maximum and minimum distances from P to the
circumference when P is without the circle.
What do the maximum and minimum values become
when P is on the circumference ?
Other simple examples are: —
MAXIMA AND MINIMA 61
(pi) The _L is the minimum distance from a given
point to a given st. line;
vj^f The minimum distance between points on two 
st. lines is _L to the  lines ;
jj^f The _l_ from a point on the circumference of a
circle to a fixed chord is a maximum when the J, or
the ± produced, passes through the centre.
60. (a) A and B are two fixed points on the same
side ofa fixed st. line CD. It is required to find the
point P in CD such that PA + PB is a minimum.
Draw BM 1 CD and pro ~*;E
duce making ME = BM. Draw ^ Q ,^' p/ A '  M
AE cutting CD at P. !*^T^ ~T^
Then P is the required \ / '"
point. i /
'/A
Take any other point Q, in
J v Fig. 66.
CD. Join PB, QA, QB, QE.
AQ + QE > AE. But QE = QB and PE = PB.
.'• AQ + QB > AP + PB ;
and hence AP + PB is the minimum value.
It is easily seen that AP and PB make equal Zs
CPA and BPD with CD, and hence: —
The sum of the distances from A and B to the st.
line is a minimum when the distances make equal Zs
with the line.
Find the point P when A and B are on opposite
sides of AB.
62
SYNTHETIC GEOMETRY
D C
x 7x .XH Y
A <0 ^B
(£)■ Of all As on the same base and having the same
area the isosceles A has the least perimeter.
Since the area is constant
the locus of the vertices is
a st. line XY  AB.
If A ACB is isosceles, and
DAB is any other A on AB
and having its vertex in
XY,
Z XCA = Z CAB = Z CBA = Z YCB ;
and .*. , by (a), AC + CB<AD+DB;
that is, the perimeter of A ACB is less than that of
any other A on the same base AB and having the
same area.
Of all As inscribed in a given acute Zd A the
al A has the least perimeter.
If DEF be any A inscribed
in ABC and FE be considered
fixed, by (a), the sum of FD
and DE will be least when
Z FDB = Z EDC; thus for the
minimum perimeter the sides
FD, DE must make equal Zs
with BC. Similarly DF, EF
must make equal Zs with AB
and DE, EF must make equal Zs with CA.
The sides of the pedal A XYZ make equal Zs with
the corresponding sides of A ABC.
.*. the perimeter of XYZ is less than that of any
other A inscribed in ABC.
MAXIMA AND MINIMA 63
(c^The rectangle contained by the two segments of
a^st. line is a maximum when the st. line is bisected.
Let P be any point in AB, and
O the middle point.
Describe the semicircle ACB
and draw OC and PD _i_ AB.
Join O, D. Fig. 58.
AP . PB = PD 2 and AO . OB = OC 2 ;
but V OC = OD and OD > PD .*. OC>PD;
and /. AO . OB > AP . PB.
(e^If the area of a rectangle is constant, its perimeter
is a minimum when the rectangle is a square.
In Fig. 58, rect. AP . PB = PD 2 .
The perimeter of the square = 4 PD while the
perimeter of the rectangle = 2 AB = 4 OD, and PD < OD.
.'. The perimeter of the square is less than that of
the rectangle of the same area.
61.— Exercises
(a)
y. Through a given point within a given circle draw the
chord of min. length.
•2L A and B are two fixed points, and CD is a fixed st.
line. Find the point P in CD, such that the difference
between PA and PB is a maximum.
(a) When A and B are on the same side of CD ;
(b) When A and B are on opposite sides of CD.
v 4fc Two sides AB, AC of a A are given in length. Show
that the area of the A is a max. when ^ A is a rt. ^.
64 SYNTHETIC GEOMETRY
4. A, B are two fixed points and P is any point. Show
that PA 2 + PB 2 is a min. when P is the middle point of
AB.
5. A, B, C are fixed points and P is any point. Show
that PA 2 + PB 2 + PC 2 is a min. when P is the centroid of
the A ABC.
(Note.— See O.H.S. Geometry, Ex. 16, Page 133.)
6. Find the max. and min. distances between two points
one on each of two given nonintersecting circles.
.7. Given two adjacent sides describe the j gm of max. area.
8. A, B are two fixed points. Find a point P on a fixed
circle such that PA 2 + PB 2 is a max. or min.
^. Of all As of given base and given vertical /, the
isosceles A has the greatest area.
10. Prove that the greatest rectangle that can be in
scribed in a given circle is a square.
11. Give examples showing that if a magnitude vary
continuously, there must be between any two equal values
of the magnitude at least one maximum or minimum value.
12. Of all chords drawn through a given point within a
circle, that which is bisected at the point cuts off the
min. area.
13. From a given point without a circle, of which O is
the centre, draw a st. line to cut the circumference in L
and M, such that the A OLM may be a max.
(Note. — Use Ex. 3 in the analysis of this problem.)
14. Given two intersecting st. lines and a point wifchin
the ^ formed by them, of all st. lines drawn through the
point and terminated in the st. lines that which is bisected
by it cuts off the min. area.
EXERCISES 65
V6. Given the base and the perimeter of a A show that
the area is a max. when the A is isosceles.
16. Of all As having a given area, the equilateral has
min. perimeter.
(Note. — Let ABC be a A having the given area and let
two of the sides AB, AC be unequal. Then, by § 60, (b),
if an isosceles A be described on BC of the same area, it
will have less perimeter. .*. if any two of the sides be
unequal the perimeter is not a min., and hence the equi
lateral A has the min. perimeter.)
17. Of all rt.Zd As on the same hypotenuse the isosceles
A has the max. perimeter.
^S^Find a point in a given st. line such that the sum of
the squares of its distances from two given points is a min.
^f. A and B are two given points on the same side of
a given st. line; find the point in the line at .which AB
subtends the max. Z.
(Note. — Describe a circle to pass through the two given
points and touch the given st. line.)
20. Two towns are on opposite sides of a canal, unequally
distant from it, and not opposite to each other. Where
must a bridge be built, _L to the sides of the canal, that
the distance between the towns, by way of the bridge,
may be a min. %
(*)
21. A  gm is inscribed in a given A by drawing from
a point in the base st. lines  to the sides. Prove that
the area of the gm is a max. when the lines are drawn
from the middle point of the base.
22. The max. rectangle inscribed in a given A equals
half the A.
(Note.— Use Ex. 21.)
66 SYNTHETIC GEOMETRY
23. One circle is wholly within another circle, and contains
the centre of the other. Find the max. and min. chords of
the outer circle which touch the inner.
24. A, B are fixed points within a given circle. Find a
point P on the circumference such that when PA, PB
produced meet the circumference at C, D respectively,
CD is a max.
(Note. — Describe a circle through A and B and touching
the given circle.)
25. Find the point in a given st. line from which the
tangent drawn to a given circle is a min.
26. Through a point of intersection A of two circles
draw the max. st. line terminated in the two circumferences.
(Note. — Draw CAD  the line of centres and any other
st. line EAF. Join C, D, E, F to the other point of inter
section and use similar /\s.)
27. P, Q, R are points in the sides MN, NL, LM of
A LMN and RQ  MN. Find the position of RQ for which
the A PQR is a max.
(Note.— Use Ex. 21.)
28. A is a fixed point within the L XOY. In OX, OY
find points C, D respectively, such that the perimeter of
the A ACD is a min.
29. A, B are points without a given circle. On the circle
find points P and Q such that L APB is a max. and L AQB
is a min.
30. The L A of the A ABC is fixed and the sum of
AB, AC is constant. Prove that BC is a min. when
AB = AC.
31. A is a fixed point within the L XOY. The st. line
BAC cuts OX, OY at B, C. Prove that BA . AC is a
min. when OB = OC.
EXERCISES 67
32. From any point D in the hypotenuse BC of a rt.^d
A ABC J_s DE, DF are drawn to AB, AC respectively.
Find the position of D for which EF is a min.
33. If the sum of the squares on two lines is given, the
sum of the lines is a max. when they are equal.
34. CAD is any st. line through a common point A of
circles CAB, DAB. Prove that CA . AD is a max. when
the tangents at C, D meet on BA produced.
(Note. — Let E, F be the centres of circles ABD, ABC
respectively. Join EA, and draw the radius FC II EA. Join
CA and produce to D. Then tangents at C, D will inter
sect on BA. Through A draw GAH terminated in the circles.
Prove GA. AH < CA . AD.)
35. Describe the maximum A DEF which is similar to a
given A ABC and has its sides EF, FD, DE passing
respectively through fixed points P, Q, R which are not
collinear.
(Note. — On QR, RP describe segments containing Zs =
Z A, Z B respectively. Through R draw a st. line  to the
line of centres of these segments and terminated in the
arcs at D, E. DQ, EP meet at F, giving the max. A DEF.
In the proof use the proposition that similar As are as the
squares on homologous sides.)
36. A, B are fixed points on the same side of a fixed
st. line XY. Place points P, Q on XY such that the
distance PQ equals a given st. line and AP + BQ is
a min.
(Note. — Through A draw AC  XY and equal to the given
length for PQ. Draw CM 1 XY and produce, making
MD = CM. Join BD cutting XY at Q. Draw AP II CQ,
cutting XY at P.)
J
1
■ M ,
68 SYNTHETIC GEOMETRY
Miscellaneous Exercises
62. — Exercises on Loci
(a)
1. Construct the locus of a point such that the _l_s from
it to two intersecting st. lines are in the ratio of two
given st. lines.
^ A fixed point O is joined to any point A on a given
st. line which does not pass through O. P is a point on
OA such that the ratio of OP to OA is constant. Find
the locus of P.
3. A fixed point O is joined to any point A on the
circumference of a given circle, P is a point on OA such
that the ratio of OP to OA is constant. Prove that the
locus of P is a circle having its centre in the st. line
joining O to the centre of the given circle.
Find the locus when P is on AO produced.
4. A fixed point O is joined to any point A on a given
st. line which does not pass through O. P is a point on
OA such that the rect. OP . OA is constant. Show that
the locus of P is a circle.
Find the locus when P is on AO produced.
,ST Through a fixed point O within an Z YXZ draw a
st. line MON, terminated in the arms of the Z, and such
that the rect. OM.ON has a given area.
Off Find the locus of a point such that the sum of the
squares on its distances from the arms of a given rt. Z is
equal to the square on a given st. # line.
7. The locus of a point, such that the sum of its distances
from two given intersecting st. lines equals a given st.
line, consists of the sides of a rectangle ; and the locus of
a point such that the difference of its distances from the
intersecting st. lines equals the given st. line, consists of
the produced parts of the sides of the rectangle.
EXERCISES ON LOCI 69
8. Given the base QR of a A and the ratio of the other
two sides, show that the locus of the vertex P is a circle
with a diameter ST in the line QR such that S, T are
harmonic conjugates with respect to Q and R.
(Note. — The circle of Apollonius, see O.H.S. Geometry,
page 235.)
(Historical Notk.— Apollonius of Perga died in Alexandria about 200 B.C.)
,.9.'AB is a fixed chord in a circle and C is any point on
the circumference. Show that the loci of the middle points
of CA, CB are two equal circles.
10. Find the locus of the points from which tangents
drawn to two concentric circles are _L to each other.
11. Construct the locus of the centre of the circle of
given radius which intercepts a chord of fixed length on a
given st. line.
12. Show that the locus of the centre of a circle of
radius R which cuts a given circle at an Z A consists of
two circles concentric with the given circle.
13. A circle rotates about a fixed point in its circum
ference. Show that the locus of the points of contact of
tangents drawn  to a fixed st. line consists of the circum
ferences of two circles.
14. AB, CD are two chords of a circle, AB being fixed
in position and CD of given length. Find the loci of the
intersections of AD, BC and of AC, BD.
15. A and B are the centres of two circles which intersect
at C ; through C a st. line is drawn terminated in the
circumferences at D and E. DA, EB are produced to meet
at P. Find the locus of P.
16. In a quadrilateral ABCD, AB is fixed in position,
AC, BC and AD are given in length : —
70 SYNTHETIC GEOMETRY
(a) Show that the locus of P, the middle point of BD
is a circle having its centre at E, the middle point of AB,
and its radius equal to half of AD ;
(b) If F is the middle point of AC, show that the locus
of the middle point of FP is a circle having its centre at
the middle point of FE and its radius equal to one fourth
of AD.
<»)
17. "What is the locus of the point P when the st. line
MN which joins the feet of the Ls PM, PN drawn to two
fixed lines OX, OY is of given length.
(Note. — Find A in OY such that AR drawn _L OX = MN.
Draw the diameter ME of the circle through O, M, N, P ; and
join NE. A MNE = AORA; ;. ME = OA. But OP = ME,
.*. OP = OA and .'. the locus of P is a circle with centre O
and radius OA.)
18. BAC is any chord passing through a fixed point A
within a given circle with centre E. Circles described on
BA, AC as chords touch the given circle internally at B, C
respectively and cut each other at D. Show that the locus
of D is a circle described on AE as diameter.
(Note. — F, G are centres of circles BAD, CAD respectively.
Join FG cutting EA at H. Prove that AFEG is a  gm
and that HD = HA.)
19. Any secant ABD is drawn from a given point A to
cut a given circle at B and D. Through A, B and A, D
respectively two circles are drawn to touch the given
circle; show that the locus of their second point of inter
section is a circle on the line joining A to the centre of
the given circle as diameter.
20. In A ABC, two circles touch AB at B and AC at
C respectively and touch each other. Find the locus of
their point of contact.
EXERCISES ON LOCI 71
(Note. — Draw BD, CD J_ respectively to BA, CA. De
scribe any circle with centre R in BD and passing through
B. Produce DC to E making CE = RB. In DC find a
point S equidistant from R and E. S is the centre of the
circle which touches AC at C and the circle with centre R
at P. Prove Z BPC = 180°  — and that consequently the
locus is a segment on BC.)
21. Any transversal cuts the sides BC, CA, AB of a
given A ABC at D, E, F respectively. The circumscribed
circles of the As AFE, CED cut again at P. Show that
the locus of P is the circumcircle of A ABC.
22. From C, any point on the arc ACB, CD is drawn
J_ ABI with centre C and radius CD a circle is described.
Tangents from A and B to this circle are produced to meet
at P. Find the locus of P.
23. Two similar As ABC, AB'C have a common vertex
A, and the A AB'C rotates in the common plane about
the point A. Show that the locus of the point of inter
section of CC and BB' is the circumscribed circle of A ABC.
24. If a A ABC remains similar to itself while it turns
in its plane about the fixed vertex A and the vertex B
describes the circumference of a circle, show that the locus
of C is a circle.
25. AB is a fixed diameter of a given circle, E the centre
and C any point on the circumference. Produce BC to D
making CD = BC. Show that the locus of the point of
intersection of AC and ED is a circle on diameter AF such
that A and F are harmonic conjugates with respect to
E and B.
26. A rectangle inscribed in a given A ABC has one of
its sides on BC. Show that the locus of the point of
intersection of its diagonals is the line joining the middle
point of BC to the middle point of the __ from A to BC.
72 SYNTHETIC GEOMETRY
(Note. — From the point where the median from A cuts
the upper side of one of the rectangles draw a 1 to BC)
27. Any chord BAC is drawn through a fixed point A
within a circle. On BC as hypotenuse a rt./d A BPC is
described such that A is the projection of P on BC. Find
the locus of P.
28. Any circle is drawn through the vertex of a given
Z. Show that the loci of the ends of that diameter which
is  to the line joining the points where the circle cuts the
arms of the / are two fixed st. lines _L to each other and
through the vertex of the given Z.
29. Through C, a point of intersection of two given
circles, a st. line ACB is drawn terminated in the circum
ferences at A and B. Prove that the locus of the middle
point of AB is a circle passing through the points of inter
section of the given circles, and having its centre at the
middle point of the st. line joining their centres.
30. From a fixed point P, two st. lines PA, PB, at rt.
Zs to each other, are drawn to cut the circumference of a
fixed circle at A and B. Show that the locus of the middle
point of AB is a circle having its centre at the middle
point of the st. line joining P to the centre of the given
circle.
31. A  gm is inscribed in a given quadrilateral ABCD
with sides  AC and BD. The locus of the point of inter
section of the diagonals of the  gm is the st. line joining
the middle points of the diagonals of the quadrilateral.
THEOREMS 73
63. — Theorems
Definition.— If A, B be the centres of two circles, and
points P, Q be found in AB and AB produced such that
AP AQ R
PB ~~ QB ~~ r
similitude of the circles.
,, the points P, Q are called the centres of
(a)
X. The centres of similitude of two circles are harmonic
conjugates with respect to the centres of the circles.
/2. Show that each of the four common tangents of two
circles passes through one of the centres of similitude of the
circles.
X If  diameters be drawn in two circles, each of the four
st. lines joining the ends of the diameters will pass through
a centre of similitude of the circles.
X. If a circle touch two fixed circles, the line joining
the points of contact passes through a centre of similitude
of the two circles.
(Note. — Use Menelaus' Theorem.)
5. The six centres of similitude of three circles lie
three by three on four st. lines.
y6; If two circles cut orthogonally, any diameter of one
which cuts the other is cut harmonically by that other.
7. In a system of coaxial circles the two limiting points
and the points in which any one circle of the system cmts
the line of centres form a harmonic range.
8. Concurrent st. lines drawn from the vertices of the
A ABC cut the opposite sides BC, CA, AB respectively at
D, E, F. Show that
sin ACF . sin BAD . sin CBE = sin FCB . sin DAC . sin EBA.
74 SYNTHETIC GEOMETRY
9. Show that the area of A ABC = j/s (s  a) (s  b) (s  c)
where 2s = a f b + c.
10. Find the area of the A ABC and also the radius of
its circuincircle, given : —
(i) a = 65 mm., b = 70 mm., c = 75 mm.;
(ii) a = 7 cm., 5=8 cm., c = 9 cm.
11. If L, M, N, be the centres of the escribed circles of
A ABC, the circumscribed circle of A ABC is the N.P.
circle of A LMN.
12. In Ex. 11 if I be the centre of the inscribed circle,
P the point where the circumscribed circle cuts IL and
PH be _L AC, AH equals half the sum and CH half the
difference of b and c.
13. If O be the orthocentre of A ABC, A, B, C, O are the
centres of the circles which touch the sides of the pedal A.
14. CA, CB are two tangents to a circle; E is the foot
of the J_ from B on the diameter AD; prove that CD
bisects BE.
(Note. — Produce DB to meet AC produced. Join AB.)
15. The _L from the vertex of the rt. L on the hypote
nuse of a rt.^d A is a harmonic mean between the
segments of the hypotenuse made by the point of contact
of the inscribed circle.
(Note. — AB is the hypotenuse of the rt.Zd A ABC,
p the length of the ± from C on AB. Then s  a, sb are
the segments of AB made by the point of contact of the
inscribed circle.
2 (s a) (sb) (b + ca)(c + ab)
H.M. of segments = ^ ^ ^= v ^ =
° sa + sb 2c
<?~a?b* + 2ab ab
2c =7 = ^>
THEOREMS 75
16. The side of a square inscribed in a A is half the
harmonic mean between the base and the ± from the
vertex to the base.
17. The circumscribed centre of a "A is the orthocentre
of the A formed by joining the middle points of its sides ;
and the two As have a common centroid.
18. ABC is a A. Describe a circle to touch AC at C
and pass through B. Describe another circle to touch BC
at B and pass through A. Let P be the second point of
intersection of these circles. Show that Z ACP = Z CBP
= Z BAP ; and that the circumscribed circle of A APC
touches BA at A. Find another point Q such that Z
QBA = Z QAC = Z QCB.
19. O is the orthocentre of A ABC, AX, BY, CZ are the
_Ls from A, B, C on the opposite sides, BD is a diameter
of the circumscribed circle. Show that : —
(a) DC = AO ;
(6) A0 2 + BC 2 = BO 2 + CA 2 = CO 2 f AB 2 = the square
on the diameter of the circumscribed circle.
20. If a A be formed with its sides equal to AD, BE,
CF, the medians of A ABC, the medians of the new A
will be respectively threefourths of the corresponding sides
of the original A.
(Note. — Draw FG  and = AD. Join CG. Produce FD,
GD to cut CG, CF at K, H.)
21. The opposite sides of a quadrilateral inscribed in a
circle are produced to meet; show that the bisectors of the
two Zs so formed are _L to each other.
22. AG is a median of the A ABC. BDEF cuts AG, AC
and the line through A  BC at D, E, F respectively.
Show that B, D, E, F is a harmonic range.
76 SYNTHETIC GEOMETRY
23. If A, C, B, D be a harmonic range, show that : —
AJL+L.
AB AC^AD
24. Prove that the "radical axis of the inscribed circle of
A ABC and the escribed circle which touches BC and
AB, AC produced bisects BC.
25. If a st. line is divided in medial section and from
the greater segment a part is cut off equal to the less,
show that the greater segment is divided in medial section.
26. If a st. line is divided in medial section, the rectangle
contained by the sum and difference of the segments is
equal to the rectangle contained by the segments.
27. In Figure 33, show that the centre of the circumcircle
of A HBC lies on the circumcircle of A AHC.
28. Show that the radius of a circle inscribed in an
equilateral A is onethird of that of any one of the escribed
circles.
29. A, B, C, D and P, Q, R, S are harmonic ranges
and AP, BQ, CR are concurrent at a point O, Prove that
DS passes through O.
30. A, B, C, D and A, E, F, G are harmonic ranges
on two st. lines AD, AG. Prove that BE, CF, DG are
concurrent.
P)
31. Three circles pass through two given points P, Q.
Two st. lines drawn from P cut the circumferences again
at R, S, T and R', S', T. Show that RS : ST = R'S' : S'T'.
(Note. — Join the six pcints to Q.)
32. The middle points of the diagonals of a complete
quadrilateral are collinear.
THEOREMS
ABCDEF is a complete quadrilateral; L, M, N the middle
points of its diagonals.
Draw LHK  AE, produce KM to cut AB at G and join
GH.
Prove L, M, N collinear.
k F
33. ABCD is a quadrilateral and O is a poi
such that A AOB + A COD = A BOC + A
that the locus of O is the st. line joining
points of the diagonals AC and BD.
(Note. — Produce DA. CB
to meet at E. Make EF =
AD and EG = BC. Join
FO, EO, GO, FG.)
A OEF = A OAD, A OEG =
A OBC. ;. OF EG  half
ABCD, and A EFG is con
stant; .'. A OFG is constant
and is on the constant base
FG ; ,\ the locus of O is a
Fio. 60.
It is easily seen that the
middle points of the diagonal are on the locus.
nt \v
AOD
the
ithin it
; show
middle
78 SYNTHETIC GEOMETRY
34. If a quadrilateral be circumscribed about a circle,
the centre of the circle is in the st. line joining the middle
points of the diagonals.
(Note.— Use Ex. 33.)
35. G is a fixed point in the base BC of the A ABC and
O is a point within the A such that A AOB + A COG =
A AOC + A BOG ; show that the locus of O is the st. line
joining the middle points of BC and AG.
(Note.— From CB cut off CD = BG. Join OD, AD.)
36. G is the point of contact of the inscribed circle of
A ABC with BC. It is required to show that the centre
of the circle is in the st. line joining the middle points of
BC and AG.
37. A is a fixed point on a given circle and P is a
variable point on the circle. Q is taken on AP produced
so that AQ : AP is constant. Show that the locus of Q is
a circle which touches the given circle at A.
38. S is a centre of similitude of two circles PQT P'Q'T',
and a variable line through S cuts the circles at the
corresponding points P, P' ; Q, Q'. Prove that, if STT' is
a common tangent
SP . SQ' = SP' . SQ = ST . ST'.
39. AD is a median of the A ABC. A st. line CLM
cuts AD at L and AB at M. Prove that ML : LC =
AM : AB.
40. The locus of the point at which two given circles
subtend equal Ls is the circle described on the join of
their centres of similitude as diameter.
41. Having the N.P. circle and one vertex of a A
given, prove that the locus of its orthocentre is a circle.
(Note. — Let A be the given vertex and K the centre
of the N.P. circle. Join AK and produce to M making
KM = AK. M is the centre of the locus.)
THEOREMS 79
42. AB is a chord of a circle and the tangents at A, B
meet at C. From any point P on the circle _Ls PX, PY,
PZ are drawn to BC, CA, AB respectively. Prove that
PX . PY = PZ 2 .
43. OX, OY are two fixed st. lines and from them equal
successive segments are cut off; AC, CE, etc., on OX; BD
DF, etc., on OY. Show that the middle points of AB, CD,
EF, etc., lie on a st. line  to the bisector of the Z XOY.
(Note. — Produce the line joining the middle points of
AB, CD to cut OX, OY and use Menelaus' Theorem.)
44. Show that the st. line joining the vertex of a A to
the point of contact of the escribed circle with the base
passes through that point of the inscribed circle which is
farthest from the base.
(Note. — Show that the vertex and the point where the
bisector of the vertical Z cuts the base are harmonic
conjugates with respect to the inscribed and escribed
centres ; and use the resulting harmonic pencil having its
vertex at the point of contact of the escribed circle with
the base.)
Prom Ex. 44 show that the _L from the vertex to the
base of a A and the bisector of the vertical Z are harmonic
conjugates with respect to the lines joining the vertex to
the points of contact of the inscribed and escribed circles
with the base.
45. The five diagonals of a regular pentagon intersect at
five points within it. Show that the area of the pentagon
7 — 3 /5~
with these points for vertices is ~ A, where A is
the area of the given pentagon.
46. ABCD is a rectangle. If A, P, C, Q and B, R, D, S
are each harmonic ranges, show that P, Q, R, S are
concyclic.
80 SYNTHETIC GEOMETRY
47. If one pair of opposite sides of a cyclic quadrilateral
when produced intersect at a fixed point, prove that the
other pair when produced intersect on a fixed st. line.
What is the connection between the fixed point and the
fixed st. line 1
48. Concurrent st. lines drawn from the vertices of the
A ABC cut the opposite sides BC, CA, AB respectively at
D, E, F. Prove that the st. lines drawn through the
middle points of BC, CA, AB respectively  to AD, BE, CF
are concurrent.
49. The sides AB, BC, CD, DA of a quadrilateral touch
a circle at E, F, G, H respectively. Show that the opposite
vertices of ABCD, the intersection of the diagonals of
EFGH, and the intersections of the opposite sides of EFGH
form two sets of collinear points.
50. The circle APQ touches the circle ABC internally at
A. The chord BC of the circle ABC is tangent to the
circle APQ at R, and the chords AB, AC intersect the
circle APQ in the points P, Q. Prove that AP . RC =
AQ . BR.
51. Tangents to the circumcircle of A ABC, at the
vertices, meet at D, E, F. AD, BE, CF are concurrent at
O. Show that the _Ls from O on the sides of A ABC are
proportional to the sides.
(Note. — Draw GDH H FE and meeting AB, AC produced
at GH. Draw DR, DS _L AG, AH. Prove DG = DH ; and,
OM DR AH AB \
if OM, ON are J. AB, AC, that o~N = DS = AG = AC°J
If AD cuts BC at K, show that BK : KC = AB 2 : AC 2 .
52. O is the middle point of a chord AB of a circle, DE,
FG are any chords through O; EF, GD cut AB at H, K.
To prove OK = OH.
THEOREMS 81
Produce EF, GD to meet at L; EG, FD at M. Produce
ED to meet LM at P. Join OL, OM.
m^:'' _ i..:E.l\ L
OLM is a selfconjugate A ; .'. CO produced cuts LM at
rt. ^s, and .*. AB il LM. PDOE is a harmonic range, and
.". L (P, D, O, E) is a harmonic pencil. Then V AB through
O is  LM and cuts the other two rays at H, K; .*. OH =
OK.
53. Through any point P in the median AD of A ABC a
st. line is drawn cutting AB, AC at Q, R. Prove that
PQ : PR = AC . AQ : AB . AR.
(Note. —Produce BP, CP to cut AC, AB at K, L. Join
3R, CQ. BD . CL . PK = DC . LP . KB,
LP PK
CL KB'
A APQ = AAPR . A APQ = A ACQ ,
•'• A ACQ AABR' ° r AAPR A ABR
PQ : PR = AC . AQ : AB . AR.)
82 SYNTHETIC GEOMETRY
64. — Problems
(a)
1. Draw a st. line, terminated in the circumferences of
two given circles, equal in length to a given st. line, and
 to a given st. line.
2. Through a given point on the circumference of a circle
draw a chord which shall be bisected by a given chord.
3. In the hypotenuse of a rt. Z d A find a point such that the
sum of the ±s on the arms of the rt. Z equals a given st. line.
What are the limits to the length of the given st. line?
4. In the hypotenuse of a rt.Zd A find a point such
that the difference of the Ls on the arms of the rt. Z
equals a given st. line.
"When will there be two, one or no solutions 1
5. In the hypotenuse of a rt.Zd A find a point such
that the ±s on the arms of the rt. Z are in a given ratio.
6. Through a given point draw a st. line terminated in
the circumferences of two given circles and divided at the
given point in a given ratio.
(Analysis. — Let A be the given point and suppose PAQ
to be the required st. line terminated at P, Q in the circles
of which the centres are respectively C, D. Join CP, and
draw QR  CP meeting CA in R. From similar As,
CA CP PA
— — = —  = the given ratio ^TTy wherein CA } CP are
known; and .*. the position of R and the length of QR
are known.)
7. In a given circle inscribe a rectangle having its
perimeter equal to a given st. line.
8. In a given circle inscribe a rectangle having the
difference between adjacent sides equal to a given st. line.
9. In a given circle inscribe a rectangle having its sides
in a given ratio.
PROBLEMS 83
10. In a circle of radius 5 cm. inscribe a rectangle having
its area 22 sq. cm.
(Analysis. — Let x, y be the sides of the rectangle, then
xy = 22 and x 2 + y 2 = 100. Show that x + y = 12.)
11. A and B are fixed points on the circumference of a
given circle. Find a point C on the circumference such
that CA, CB intercept a given length on a fixed chord.
(Analysis. — Draw BL_  to the given chord and equal to
the given length. Join L to the point where AC is supposed
to cut the given chord. Join AL, etc.)
12. A and B are fixed points on a circumference. Find
a point C on the circumference such that CA, CB cut a
fixed diameter at points equally distant from the centre.
(Analysis. — Draw the diameter AOD. Join D to the
point F where BC is supposed to cut the given diameter.
Prove that the circumcircle of A DFB touches the given st.
line AD at D.)
13. In a given circle inscribe a A, such that two of its sides
pass through given points, and the third side is a maximum.
14. Two towns are on different sides of a straight canal,
at unequal distances from it, and not opposite to each
other. Where must a bridge be built _L to the direction
of the canal so that the towns may be equally distant from
the bridge 1
15. Divide a given st. line into two parts so that the
squares on the two parts are in the ratio of two given st.
lines.
16. Construct the locus of a point the difference of the
squares of whose distances from two points 3 inches apart
is 5 sq. inches.
17. Two points A and B are four inches apart. Con
struct the locus of the point the sum of the squares of
whose distances from A and B is 20'5 square inches.
84 SYNTHETIC GEOMETRY
18. Divide a given st. line into two parts such that the
sum of the squares on the whole st. line and on one part
is twice the square on the other part.
(Analysis.— Let AB (= a)
Ct^.^^ be the required line and
N v x "*«. EB (= cc) a segment such that
iA %N y/ ""*V^. B " 2 + & = 2(a  x)\ Then
D E cc = 2a  aj/3, and a  x —
Fig. 62. ,5 . AE \/3  1 _
ay8 a; .. ^ = ^VS 
1/3 + 1. .'. AE = EB /3+ EB. Cut off ED = EB and draw
AC J_ AB and = EB. Then since AD = EB ]/3 = AC v/3,
Z ADC = 30° and CD = 2AC = DB. .'. Z B = 15°.
The following construction is then evident : — At B make
Z ABC = 15°, and draw AC __ AB. Draw the rt. bisector
of BC cutting AB at D. Bisect DB at E.)
19. Two nonintersecting circles have their centres at A
and B, and C is a point in AB. Draw a circle through
the point C and coaxial with the two given circles.
(Note. — From O, the point where the radical axis cuts
AB draw a tangent, OT, to one of the given circles. Then
if CC is the diameter of the required circle, OC . OC =
OT 2 and .*. OC is the third proportional to OC and OT.)
20. Construct a A having one side and two medians
equal to three given st. lines. (Two cases.)
21. Construct a A having the three medians equal to
three given st. lines.
22. Given the vertical Z, the ratio of the sides containing
it, and the diameter of the circumscribing circle; construct
the A.
23. Given the feet of the Ls drawn from the vertices of
a A to the opposite sides; construct the A.
PROBLEMS
85
24. Draw a circle to touch a given circle, and also to
touch a given st. line at a given point.
25. Draw a circle to pass through two given points and
touch a given circle.
26. Draw a circle to pass through a given point and
touch two given intersecting st. lines.
27. AB is the chord of a given segment of a circle.
Find a point P on the arc such that AP + BP is a
maximum.
28. Find a point O, within a A ABC such that: —
(1) A AOB : A BOC : A COA 1:2:3;
(2) A AOB : A BOC : A COA = I : m : n.
(b)
29. Find a point such that its distances from the three
sides of a A may be proportional to three given st. lines.
(Note. — Draw BD, CE J_
BC, and each = I. Join DE.
Draw AF ± AC, and = m.
Draw FG II AC, meeting DE
at G. Draw AH _L AB, and
= n. Draw HK  AB, meet
ing DE at K. BK, CG meet
at P the required point.
Show that, if PL, PM, PN
be ± to the sides, PL : PM
30. Through a given point within a circle draw a chord
which shall be divided in a given ratio at the given point.
31. A, B, C, D are points in a st. line. Find a point
at which AB, BC, CD subtend equal ^s.
(Note. — The required point is the intersection ot two
circles of Apollonius. See O.H.S. Geometry, page 235.)
86
SYNTHETIC GEOMETRY
32. Given a vertex, the orthocentre and the centre of
the N.P. circle of a A, construct the A.
33. Having divided a st. line internally in medial section,
find the point of external division in medial section hy ratio
and proportion.
3i. Describe an equilateral A with one vertex at a given
point, and the other two vertices on two given  st. lines.
(Analysis. — Describe a
circle about the equilateral
A ABC cutting the  lines
again at L, M. Then
Z ALC = ABC = G0°, and
Z AMB = Z ACB = 60°.)
35. Find the locus of
the middle point of the
chord of contact of tan
gents drawn from a point
on a given st. line to a
Fia 64  given circle.
36. The locus of the centre of a circle which bisects the
circumferences of two given circles is a st. line _l_ to the
line of centres and at the same distance from the centre of
one circle that the radical axis is from the centre of the other.
37. Describe a circle to bisect the circumferences of three
given circles.
38. Find the locus of the centre of a circle that passes
through a given point and also bisects the circumference of
a given circle.
39. Describe a circle to pass through two given points
and bisect the circumference of a given circle.
40. Given a point and a st. line, construct the circle to
which they are pole and polar, and which passes through a
given point.
Part II.— ANALYTICAL GEOMETRY
FORMULA
The following important results from algebra and trigonometry
are frequently used in analytical geometry : —
1. The roots of the quadratic equations ax 2 + 2bx + c = ara
 6 + jW~^ and  b  ,jW^ t
a a
These roots are
real, if b 2 >, or =, aa ;
imaginary, if b 2 < ac ;
equal to each other, if 6 2 = ac ;
equal in magnitude but
ODDosite in sign, if b = ;
xationai, if b  ac is a perfect square.
One root = 0, if c = ;
ooth roots = 0, if b = c = 0.
The sum of the roots = —
a
The product of the roots = — .
o
2. The fraction ^ = oo, if 6 = and a is not = 0.
o
3. The equation ax + by + c = is the same as px + qy + r = 0,
., a b c
if — = _ = _.
p q r
4. ax 2 + 2bx + c is a perfect square, if b 2 = ac.
5. If a x x + b x y + c,_z =
and a 2 x + b 2 y + c 2 z = 0,
then * = t.
b l c 2  b 2 c 1 c 1 a 2  c 2 a 1 a 1 6 2 — a 2 b r
V F0RMULJ5
6. For all values of «,
sin 2 a + cos 2 a = 1.
7. If tan a = k, a = tan ~ l k.
8. sin (A ± B) = sin A cos B ± cos A sin B.
cos (A ± B) = cos A cos 3 + sin A sin B.
9. sin A + smB = 2 sin ~*~ , cos ^1 — , etc
in i tK j. q\ ton A ± tan B
10. ton (A ± B) = _— — .
1 + tan A ton B
CONTENTS
Chapter I ?Aea
Cartesian Coordinates , 1
Rectangular Coordinates „ . 2
The Distance Between Two Points 6
Area of a Triangle 14
Loci 18
Chapter II
The Straight Line 25
The Angle Between Two Straight Lines ... 40
Perpendiculars ... 45
Chapter III
The Straight Line Continued 57
Transformation of Coordinates 62
Review Exercises . . .' 68
Chapter IV
The Circle . . . 72
Tangents 79
Poles and Polars 88
Tangents from an Outside Point ..... 95
Radical Axis , . . 97
Miscellaneous Exercises 99
Answers . . ......... 113
ELEMENTARY ANALYTICAL GEOMETRY
CHAPTER I
Cartesian Coordinates
1. Analytical, or algebraic, geometry was invented
by Descartes in 1637, and this invention marks the
beginning of the history of the modern period of
mathematics. It differs from pure geometry in that
it lays down a general method, in which, by a few
simple rules, any property can be at once proved or
disproved, while in the latter each problem requires a
special method of its own.
2. The Origin. In plane analytical geometry the
positions of all points in the plane are determined by
their distances and directions as measured from a fixed
point.
If the points are all in a st. line, the fixed point is
most conveniently taken in that line.
if— 3 c 1
Fig. 1.
Thus, if the distance and direction of each of the
points A, B, C, D, E from the point O are given, the
positions of these points are known.
The point O is called the origin, or pole.
1
2 ELEMENTARY ANALYTICAL GEOMETRY
3. Use of plus and minus. In algebra the signs
plus and minus are used to indicate opposite qualities
of the numbers to which they are prefixed ; and in
analytical geometry, as in trigonometry, these signs are
used to show difference of direction. In a horizontal
st. line distances measured from the origin to the
right are taken to be positive, while those to the left
are negative ; and in a vertical st. line distances
measured upward are positive, while those measured
downward are negative. Thus, in Fig. 1, if OA = 2 cm.,
OB = 3 cm., OC = 5 cm., OD = 1 cm., and OE = 3 cm.,
the positions of these points are respectively repre
sented by 2, 3, 5, — 1 and — 3, the understood unit
being one centimetre.
Rectangular Coordinates
4. Coordinates. When points are not in the same
st. line, their positions are determined by their distances
from two st. lines xOx and y'Oy drawn through the
origin, the distances being measured in directions  to
the given st. lines.
RECTANGULAR COORDINATES 6
These lines are called the axes of coordinates, or
shortly, the axes.
x'Ox is called the axis of x, and y'Oy is called the
axis of y.
From a point P draw PM i Oy and PN  Ox, terminated
in the axes.
PM is called the ordinate of P, and PN ( = OM), is
called the abscissa of P. These two distances, the
abscissa and ordinate, are called the coordinates of
the point.
Sometimes, from the name of the inventor, they are
spoken of as cartesian coordinates.
5. Rectangular coordinates. When the axes are at
rt. Zs to each other, the distances of a point from
the axes are called its rectangular coordinates.
T P
I I I
Fia. 3.
To locate the point of which the abscissa is 4 and
the ordinate 3 when the coordinates are rectangular,
4 ELEMENTARY ANALYTICAL GEOMETRY
measure the distance OM = 4 units along Ox and at
M erect the J_ PM = 3 units. P is the required point.
f!
1
/
t\
/\
I
1
''
1
Fig. 4. (Unit = J inch. )
In Fig. 4, the abscissa of P = OM = 2 8, the ordinate
of P = PM = 2. The position of this point is then
indicated by the notation (2*8, 2). For Q, the abscissa =
ON =  1'6, the ordinate = QN = 26 and the position
of the point is indicated by ( — 1*6, 2  6).
Similarly the position of R is ( — 1, — 1'6), and
that of S is (14, 12).
xOy, yOx', x'Oy' and y'Ox are respectively called
the first, second, third and fourth quadrants; and we
see from the diagram, that: —
for a point in the first quadrant both coordinates
are positive;
EXERCISES 5
for a point in the second quadrant the abscissa is
negative and the ordinate is positive ;
for a point in the third quadrant both are negative;
and
for a point in the fourth the abscissa is positive and
the ordinate is negative.
Thus the signs of the coordinates show at once in
which quadrant the point is located.
6.— Exercises
A^ Write down the coordinates of the points A, B, C, D,
E, F, G, H and O in Fig. 5.
f t\
~?
^i\r
\l)
1
.
*>
*
..
■s
/ \ i
\>
■F
(Z
Fig. 5. (Unit = ^ inch.)
' J£. Draw a diagram on squared paper, and mark on it
the following points :— A (4, 3), B (46, 0), C (  2,  3),
D (  4, 2), E (0, 28). Indicate the unit of measurement
on the diagram.
&? Draw a diagram on squared paper and mark the
following points :— (4, 3), (3, 4), (  3,  4), (  4, 3), (0, 5),
(0,  5), (  5, 0). Describe a circle with centre O and
6
ELEMENTARY ANALYTICAL GEOMETRY
radius 5. Should the circle pass through the seven points'!
Why?
4. The side of an equilateral A = 2a. One vertex is at
the origin, one side is on the axis of x and the A is in the
first quadrant. What are the coordinates of the three
vertices 1
^ 5. One corner of a square is taken as origin and the axes
coincide with two sides. The length of a side is b. What
are the coordinates of the corners, the square being in the
first quadrant 1 ?
The Distance Between Two Points
7. In general, the abscissa of a point is represented
by x, the ordinate by y.
(C)To find the distance between a point P (x v yj
and the origin.
y'
Fiq. 6.
From P draw PM j_ Ox.
V PMO is a rt.Zd A,
.. PO 2 = OM ! f PM 2
/. po = ^ + y x 2 .
THE DISTANCE BETWEEN TWO POINTS
(9JT0 find the distance between P (x v y x ) and
Draw PM and QN j_ Ox ; QL j_ PM.
QL = NM = OM  ON = x l  X.,.
PL = PM  LM = PM  QN = y l  y 2 .
V PLQ is a rt.Zd A,
.*. PQ 2 = QL 2 + PL 2 .
= (x l  x 2 y + ( Vl  yj*.
.. PQ = v /( Xl  x. 2 y + ( 7l  y 2 )l
10. If the point Q in § 9 coincides with the origin
O, x., = and y. 2 = 0. Substituting these values, in the
expression for PQ in that article we obtain
PO = yx* + y x \
This shows that the result in §8 is a particular case
of that in § 9.
11. The result in § 9 holds good, in the same form,
for any two points whether the coordinates are positive
or negative.
8
ELEMENTARY ANALYTICAL GEOMETRY
For example — it is required to find the distance
between P ( 3, 2) and Q (5,  2).
*N
N s
k
■
^s
s s
<>
Fia. 8. (Unit = ^ inch.)
Draw PM, QN j. Ox ; QL 1 PM.
The length of ML = length of NQ = 2.
.. PL = PM + ML = 2 + 2 = 4.
The length of QL = NM = 5 + 3 = 8.
PQ^ = ql 2 + PL 2
= 64+ 16 = 80.
.'. PQ = 4 y'W.
If in the expression for PQ found in § 9, we sub
stitute  3 for x v 2 for y v 5 for x 2 and  2 for y 2 ,
we obtain
PQ. i/(35)2 + (2 + 2)s
= 4j/5,
the same result.
THE DISTANCE BETWEEN TWO POINTS 9
ff2j)The particular cases in § § 10 and 11 illustrate
what is known as the continuity of the formulae
in analytical geometry. Here continuity means, that
general results which are obtained when the coordinates
in the diagram used are all positive hold true in the
same form for all points.
13. To find the coordinates of the middle point
of the distance between two given points P (x v y x )
and Q (x 2 , y 2 ).
y
Qf*
JpC_JT
1 i
—  js 1
1 1
N
L M
X
Let R (x, y) be the middle point of PQ.
Draw PM, QN, RL ± Ox ; QS j_ RL; RT 1 PM.
From the equality of As PRT, RQS,
QS = RT and RS = PT.
.*. NL = LM,
• • \JU ~~ \fj(y —— JOt JOm
x x 4 x. 2
x =
V RS = PT,
.*. yv*
Vx  y
y =
2/i + y 2
10 ELEMENTARY ANALYTICAL GEOMETRY
Thus the coordinates of R are
*i + x 2 7i + y 2
2 ' 2 '
Qj) To find the coordinates of the point dividing
the distance between P (x 1 y x ) and Q (.r 2 y 2 ) in the
ratio of m to n.
Fia. 10.
Let R (x, y) be the point dividing PQ such that
PR m
RQ — n
Draw PM, QN, RL J. Ox ; QS J_ RL; RT ± PM.
From the similar As PRT, RQS
RT _ PT P R m
QS  RS RQ ■
RT
QS
a? — cc„
m
n. '
?n
71
mx.
EXERCISES
li
.".
nx, + mx 9
x —
m + n
V
PT_ m
RS ,7'
•••
2/12/ m
2/  2/2 »
my  my. 2 = ny 1
y = ^2/i + my 2
m + 7i
 ^y.
Thus the coord
mates of R are
tj + mx. 2 ny! + my 2
f Tx*» frTv^flVu
113
>v /U."
m + n m + n
15. If the point R be taken in PQ produced such that
PR : RQ = m : n, and the coordinates of P, Q be
( x v 2/i)' ( x 2> 2/2) ik may be shown by a proof similar to
that in the previous article that the coordinates of R are
mx 2  nXj my 2  ny l
m  n ' m  n
These results and also those of § § 13 and 14 are
the same for oblique and rectangular axes.
16.— Exercises
1. Find the distance between the points (G, 5) and (1,  7)
and test your result by measurement on squared paper.
' 2. Find the distance between the points (2,  3) and
(1, 1) and test your result by measurement on squared
paper.
3. Find the coordinates of the middle points of the st. lines
joining the pairs of points in exercises 1 and 2 respectively
and test the results by measurements on the diagrams.
12 ELEMENTARY ANALYTICAL GEOMETRY
4. Find, to two decimal places, the distance between
(  3, 7) and (4,  4).
5. The vertices of a A are (  2, 4), (  8,  4) and (7, 4).
Find the lengths of its sides.
6. The vertices of a A are (  1, 5), (  4,  2), (5,  3).
Find (a) the lengths of the sides ; (b) the lengths of the
medians.
7. The vertices of a quadrilateral are (4, 3), (  5, 2),
(  3,  4), (6,  2). Find the lengths of its sides, and also
of its diagonals.
v 8. Find the coordinates of the middle point of the st.
line joining (3,  2) and (  3, 2).
9. Find the points of trisection of the st. line joining
(1, 3) and (6, 1).
10. The st. line joining P (  4,  3) and Q (6,  1)
is divided at R (x, y) so that PR : RQ =5:2. Show that
x = 2y.
11. Find the length of the st. line joining the origin to
(a,  b).
12. The st. line joining the origin to P (4, 7) is
divided at R, Q so that OR : RQ : QP = 3 : 4 : 2. Find
the distance RQ.
13. The length of a st. line is 17 and the coordinates of
one end are (  5,  8). If the ordinate of the other end
is 7, find its abscissa.
J 14. Find in its simplest form the equation which expresses
the fact that (x, y) is equidistant from (5, 2) and (3, 7).
15. Find the centre and radius of the circle which passes
through (5, 2), (3, 7) and (  2, 4).
EXERCISES 13
]%>. Find the points which are distant 15 from (2,  10)
and 13 from (2, 14).
v ^7. Prove that the vertices of a rt.Zd A are equidistant
from the middle point of the hypotenuse.
Suggestion : — Take the vertex of the rt. Z for origin and
the sides which contain the rt. Z for axes.
 ISC' In any A ABC prove that
AB 2 + AC 2 = 2 (AD 2 f DC 2 ),
where D is the middle point of BC.
Suggestion : — Take D as origin, DC as axis of x and the
J_ to BC at D as axis of y. Let DC = a, and the coordi
nates of A be (x v y x ).
19. If D is a point in the base BC of a A ABC such
that BD : DC = m : n, show that
n AB J + m AC 2 = (m + n) AD 2 + n BD 2 + m DC 2 .
Suggestion : — Take D as origin, DC as axis of x and the
J_ to BC at D as axis of y. Let BD =  ma, DC = na, and
the coordinates of A. be (x v y^.
20. The vertices of a A are the points (x v y^), (x 2 , y 2 ),
(tc 3 , y 3 ). Find the coordinates of its centroid.
v 21. The st. line joining A (2, 1) to B (5, 9) is produced
to C so that AC : BC = 7:2. Find the coordinates of C.
22. The st. line joining A (3,  2) to B (  4,  6) is
produced to C so that AC : BC = 3 : 2. Find the coordi
natas of C.
14
ELEMENTARY ANALYTICAL GEOMETRY
The Area of a Triangle
17. To find the area of the A of which the
vertices are A (x v y^ B (.»■.,, y 2 ) and c (.r 3 , y. 6 ).
Fio. 11.
Draw the ordinates AL, BM, CN.
From the diagram,
A ABC = ALNC + CNMB  ALMB.
The area of a quadrilateral of which two sides
are  = half the sum of the  sides X the distance
between the j sides.
.'. ALNC = £ (AL + CN) X LN = \ (y, + y s ) (x s  x x ),
CNMB = h (CN + BM) XNM = (i/ 3 4 2/ 2 ) ( x 2 ~ x z)>
ALMB = h (AL 4 BM) X LM = h (y 1 + 2/o) 0' 2  X x ).
;. A ABC = i { (y 1 4 y z ) (x 3  x,) + (2/3 + y 2 ) (x. 2  x z ) 
(3/1 + 2/2) (®2 «l)}
Simplifying,
A ABC = £ { Xl (y 2  y s ) 4 x, (y s  y x ) 4 x 3 (y x  y 2 )} •
THE AREA OF A TRIANGLE
15
Note. — The points have been taken in circular order about
the A in the opposite direction to that in which the hands of
a clock rotate; if they are taken in the same direction as the
hands rotate, the formula will give the same result only it
will appear to be negative ; but, of course, the area of a A
must be positive.
18. To find the area of the A of which the vertices are
(3, 2), (4,3), (2, 4).
Fig. 12. (Unit = , 3 „ inch.)
Draw the diagram on squared paper. Draw the
ordinates AL, BM, CN. Through C draw RCS  Ox to
meet AL, BM produced at S, R.
= BRSA
91
2
A ABC
BRSA
A BRC  A ACS.
l(BR + AS)RS=i(7 + 6)X7
16 ELEMENTARY ANALYTICAL GEOMETRY
14
ABRC = BRXRC = X7X2 =— •
u
30
AASC =ASXSC = £X6X5 = —
, A ABC = 911430 = 47
^ 2 2
If we substitute the coordinates of A, B and C in the
formula of § 17, we obtain
AABC= i {3(3 + 4) + (4)(4 2) + ( 2)(23)}
47
= 1(21 + 24 + 2)=^;
the same result as before.
This illustrates the continuity of the symmetrical
result found in § 17 for the area of a A.
19. — Exercises
v Q) Find, from a diagram, the area of the A of which
the vertices are (o, o), (a, b), (c, d). Check your result by
using the formula of § 17.
2. Draw the following As on squared paper and find
their areas; checking your results by using the formula of
§ 17:—
J& (1,4), (2,2), (5, 1);
(b) (4, 2), (5,1), (2, 6);
(c) (0,0), (3,45), (25,4).
3. Find the area of the quadrilateral of which thtf
vertices are (3, 6), (  2, 4), (2,  2) and (7, 3).
4. Find the area of the quadrilateral of which the
vertices are (0, 0\, (4, 0), (3, 6) and (  3, 3).
. THE AREA OF A TRIANGLE 17
^ 5. D, E, F are respectively the middle points of the sides
BC, CA, AB of a A. Prove by the formula of § 17,
taking B as origin and BC as axis of x, that A ABC  4 A
DEF.
^NL Find the area of the A of which the vertices are
(x, y), (3, 5), (2, 4); and thence show that if these
points are in a st. line 5y  x = 22.
&L Find the area of the A A (  3, 2), B (7, 2), C (3, 10);
and show that the J_ from A to BC = BC.
V 'S^ A man starts from O and goes to A, from A to B,
B to C, C to D, D to O. If O be taken as the origin
and the coordinates of A, B, C, D are (0,  3), (8, 3),
(4, 8), (4, 3), find the distance he has travelled, the
unit being one mile.
'^^ Show from the formula for the area of a A that
A (3, 2), B (19, 10) and C (7, 1) are in the same st.
line. Find the ratio of AC to CB.
10. Show that if the coordinates of the vertices taken in
order of a quadrilateral are (x v y 1 ), (x.,, y.^), (x v y.^j and
( x v 2/4). its area is
\ { x i (y 2  vd + x 2 (y s  vi) + *» (y*  vd +  r 4 Oa  ^}
^14. In the A OAB, P is taken in OA, Q in AB and R
in BO so that OP : PA = AQ : QB = BR : RO = 3 : 1.
Show that A PQR : A OAB = 7 : 16.
18
ELEMENTARY ANALYTICAL GEOMETRY
Loci
20. The definition of a locus (see Ontario H. S.
Geometry, page 77) is : —
When a figure consisting of a line or lines con
tains all the points that satisfy a given condition,
and no others, this figure is called the locus of
these points.
The condition which the points satisfy may be
expressed in the form of an equation involving the
coordinates of the points. For example, take the locus
of the points of which the ordinate is equal to 3.
This condition, which is expressed by the equation
y = 3 [i.e.: — Ox + y = 3], is satisfied by an infinite
number of points, as (0, 3), (1, 3), (2, 3), (7, 3), (4, 3),
etc. All such points are on a st. line AB  to Ox and 3
y
■8
Fig. 13.
units above it ; and this st. line contains no points which
do not satisfy the condition. Thus the equation y = 3
represents the line AB.
LOCI
19
Similarly the equation y = — 3 represents a st. line
 Ox and three units below it ; x = 3 represents a st.
line  Oy and three units to the right of the origin,
and x — — 5 a st. line  Oy and 5 units to the left
of the origin.
For another example let us take the condition to be
that the abscissa and ordinate of each point are equal.
The points (0, 0), (1, 1), (2, 2), (4, 4), (.1, 1),
( — 5, —5), etc., satisfy this condition. It is expressed
by the equation y = x. If we draw a diagram on
H X
_l2,tk? _
H * T 5?^AT
7
^
t*.ikZ
7
2
* y
~" y'
7
. 7
ZH^r
7
7
TC7
232^ ^ r
Fio. 14. (Unit = A inch.)
squared paper, mark some of these points on it and
join them we get a st. line AB bisecting the Ls xOy
and x'Oy' every point on which satisfies the given con
dition. Between O and (1, 1) there are an infinite
number of points, (h, h), (J, £), ( T V, T V), ( T V, T V), etc.,
which satisfy the condition, and so on continuously
throughout the line. Thus the equation y = x represents
the line AB.
20
ELEMENTARY ANALYTICAL GEOMETRY
Again, we may consider the point which moves so
that its distance from the origin is always 5. Its locus
is plainly the circumference of a circle. Particular
1 1
" (0.5)
J3.4)y^ ^S,d 3 ^)
/ V( 4  3 )
/ s
~7 C
f \
' 1 'I
t5.0) \S&) '
*._ _£ _
: t _l jt
x t
X t
\ ^ t
L V J
\ Z
± 5 z
^s — iosv^
: z *Jt ::: ::::
Fig. 15. (Unit = ^ inch.)
points on this locus are (5, 0), (4, 3), (3, 4), (0, 5),
( — 3, 4), etc., and its equation is Jx 2 + 2/ 2 = 5, or
+ 2/ 2
25.
21. In the equation of a locus the numbers that
are the same for all points on the locus are called
constants ; while those that change in value con
tinuously from point to point are called variables.
Thus, in the equation x 1 + y 2 = 25, x and y are
variables and 25 is a constant.
EXERCISES 21
22.— Exercises
sJl. Find four or five points on the iOcus represented by
each of the following equations ; and draw the locus on
squared paper in each case : —
(a) x =  4 ; (6) x + y = ; (c) x  2y = ;
ifi 3x + y = j (e) x = y+i; (/) ^ + ^ = 169.
v ^2^ A point moves so that its distance from the axis of
a: is 5 times its distance from the axis of y. Find the
equation of its locus.
v Jk What locus is represented by the equation (a) y = ;
(6) x = 0?
M^A point moves so that it is equidistant from the
origin and from (8, 0). Find the equation of its locus.
v/jx A point moves so that it is equidistant from the
origin and from (3,  5). Find the equation of its locus,
and draw the locus on squared paper.
v<6r A point is equidistant from (1, 2) and ( — 3, 4).
Find the equation and draw the locus on squared paper.
7. A point moves so that its distance from (4, 3) is
always 5. Find the equation and show that the locus
passes through the origin.
* 8. The coordinates of the ends of the base of a A are
(2, 3) and (4, 1), and the length of the median
drawn to the base is 6. Find the equation of the locus of
its vertex.
9. The coordinates of the ends of the base of a A are
(0, 0) and (5, 0), and its area is 10. Show that the
equation of the locus of its vertex is y = 4.
V10. The coordinates of the ends of the base of a A are
(  1, 2) and (5, 1) and its area is 9. Find the equation
of the locus of its vertex.
22 ELEMENTARY ANALYTICAL GEOMETRY
23. An equation connecting two variables x and y
has an infinite number of solutions. For example, in
the equation y = 3 x + 7, if any value is given to x,
the corresponding value of y may then be determined.
Thus, when
(a) x =
0,2/ =
7,
(b)x =
1,2/ =
10,
(c) X =
2,2/ =
13,
(d)x =
1,2/ =
4,
(e) x =
3,2/ =
2,
(f)x =
hy =
8,
etc.
The, in general, continuous line which passes through
all the points (a), (6), (c), etc., is the locus represented by
this equation.
Another equation as 4>x + 3y = 8 has also an infinite
number of solutions, and if these two equations are
solved together, the common solution obtained, in this
case x = —1, y = i, gives the coordinates of the point
of intersection of the loci represented by the equations.
Sets of solutions which satisfy the equation
4cc + Sy ;= 8 are given in the following table :— "■
(cZ)l
{S) 2
(A) 5
y
EXERCISES 23
If we plot these two sets of results on squared
Fig. 16. (Unit = & Inch.)
paper, we see that the loci appear to be st. lines
which intersect at the point (d) ( — 1, 4).
24.— Exercises
X. Plot the following loci on squared paper and find the
coordinates of their points of intersection : —
tyrf 4a;  y >= 1 and x  2y =  12
(b) x + 2y = 7 and 5x  2y = 11
(c) 3,r + 8y =  18 and 4r + 3y =  1
(cl) 3.c f 4y = and x 2 + y 2 = 100
(e) 3a:  5y f 45 = and x 2 + y 2 = 169.
V2f Find the points where the locus 3x  5y f 45 =
cuts the axes.
3. Find the points where the locus x 2 + y' 2 = 6 x cuts
the axis of x.
24 ELEMENTARY ANALYTICAL GEOMETRY
4. Find the locus of a points such that the square of its
distance from (  a, o) is greater than the square of its
distance from (a, 6) by 2 or.
•5. Find the equation of the locus of a point such that
the square of its distance from (2,  1) is greater than
the square of its distance from (5, 3) by 11.
6. A (1, 0) and B (9, 0) are two fixed points and P is a
variable point such that PB = 3 PA. Find the equation of
the locus of P.
7. Plot the following loci and show that they are
concurrent : —
%x + 4y = 10, 5x  2y = 8, 4x f y  9.
CHAPTER II
The Straight Line
(25. To find the equation of a st. line in terms of
the intercepts that it makes on the axes.
Let the st. line cut the axes at A, B so that OA
OB = b.
a,
Take P (x, y) any point on the line, and draw PM \\Oy
and terminated in Ox at M.
From the similar As APM, ABO,
PM AM
BO ~ AO "
y a—x,
b a
a + b L
Note. — It is seen from the diagrams that both the proof
and the form of the equation are the same for oblique and
rectangidar axes.
25
26
ELEMENTARY ANALYTICAL GEOMETRY
2G. To find the equation of the st line passing
through A (x v y x ) and B (a^ y 2 ).
P.
Take any point P (x, y) on the st. line.
Draw AK, BL PM  Oy and terminated in Ox at
K, L, M ; and AN, BR  Ox and respectively terminated
in PM at N and AK at R.
From the similar As PNA, ARB,
AN
BR
PN
AR'
AN = KM = OM  OK = X  X v
BR = LK = OK  OL = x 1  X 2 ,
PN = PM  NM = PM  AK = y  y v
AR  AK  RK = AK  BL = y x  y 2
. x  x t _ y  y t
x L x 2 y!y 2 '
Note. — It is seen from, the diagrams that both the proof
and the form of the equation are the same for oblique and
rectangular axes.
EXERCISES 27
27. — Exercises
1. The equation of the st. line passing through (4, 3) and
(2, 7) is by the formula of § 26
x  4 y  3
4+2 = 3^7 <*
or, 2x + 3y = 17.
To find the intercepts which this line makes on the axes,
let y = and .*. x = 8£, let a; = and .*. y = 5f. By § 25
the equation of the line may now be written
8i + 5f
This is clearly the same as 2x f Zy = 17.
2. Write down the equations of the st. lines which make
the following intercepts on Ox, Oy respectively : —
(a) 5, 2; (b) 4, 6; (c) 3, 8.
3. Find the equations of the st. lines through the follow
ing pairs of points : —
{a) (6, 2), (3, 1); (b) (  1, 2), (3, 7); (c) (4, 6),
(  7, 2). Find the intercepts these st. lines make on the
axes.
4. Find the point where the st. line which makes inter
cepts  3 and 5 on Ox and Oy respectively is cut by the
st. line x =  5.
">. Find the point where the st. line making intercepts 7
and 2 on Ox and Oy respectively meets the st. line through
(  2, 7) and (5,  3).
6. Find the point where the st. line through (3, 5) and
(7,  1) meets the st. line through (8, 2) and (6, 5).
28 ELEMENTARY ANALYTICAL GEOMETRY
1 7. Prove that (11, 4) lies on the st. line joining (3, 2)
and (19, 10) and find the ratio of the segments into which
the first point divides the join of the other two.
8. Find the equations of the sides of the A of which
the vertices are (4, 2), (5, 1), and (2, 6). Find
also the equations of the medians of the A and the coordi
nates of its centroid.
9. The vertices of a quadrilateral are (3, 6), (  2, 4),
(2,  2) and (7, 3). Find the equations of the four sides.
Find also the equations of the three diagonals of the
complete quadrilateral, and show that the middle points of
the diagonals are collinear. Find the equation of the st.
line passing through the middle points of the diagonals.
10. Find the vertices of the A the sides of which are
11.x  3y =  45, 5x  lly = 47 and 3x + ly = 7.
11. P (ajj, y x ) is any point and  + '' = 1 cuts Ox, Oy at
A, B respectively. Show that the area of the A PAB =
l (bx x + ay x  ab).
THE STRAIGHT LINE 29
28. As explained in elementary algebra, the degree
of a term, with respect to certain letters, is the number
of such letters that occur as factors in the term.
Sx, — by, ax, by are terms of the first degree with
respect to x and y.
5x 2 , Sy 2 , —Ixy, ax 1 are terms of the second degree
with respect to x and y.
29. Degree of an equation. An equation is said to
be of the first degree in x and y when it contains a
term, or terms, of the first degree in x and y, but no
term of a higher degree than the first.
The general equation of the first degree in x and
V is
Ax + By + C = 0.
An equation is said to be of the second degree in
x and y when it contains a term, or terms, of the
second degree in x and y, but no term of a higher
degree than the second.
The general equation of the second degree in x
and y is
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0,
or, in a more convenient form,
ax 2 + 2hxy + by 2 + 2gx + 2fy f c = 0.
30 ELEMENTARY ANALYTICAL GEOMETRY
30. To prove that an equation of the first degree
always represents a st. line.
Let (x lt 2/1), (#2> 2/>)> ( x z> 2/3) De an y three sets of
simultaneous values of x and y which satisfy the
equation Ax \ By + C = 0.
Then, (1) Ax X + By 1 j C = 0.
(2) Ax 2 +By 2 + C = 0.
(3) Ax s + B7/ 3 + C = 0.
From (1) and (2),
A B C
2/l  V2 X 2 ~ X l X lV 2  X 2Vl
Dividing the three terms of (3) respectively by
these equal fractions and by any one of them,
x s (Vi ~ 2/2) + 2/3 O2 ~ x i) + x i V2  » 2 Vx = °
Rearranging the terms, we get
( 4 ) x i (2/2 ~ 2/3) + ® 2 (2/3 ~ 2/i) + »3 (2/i  2/2) = 0
From § 17 the area of the A formed by joining
(x v y x ), (x 2 , y 2 ), (x 3 , 2/3) is
I { x i (2/2 ~ 2/3) + X 2 (2/3 ~ 2/i) + *3 (2/i ~ 2/2)}
and .'., from (4), in this case, the area of the A is
zero.
This can only be so when the three points are in a
st. line, and .'. as any three points the coordinates of
which satisfy
Ax + By f C =
are in a st. line, this equation must always represent
a at. line.
THE STRAIGHT LINE 31
(3Lj The equation Ax {By f C = may be changed
to the form
A_ + £=!,
C _ C
A B
and by comparing this with the equation of § 25,
x y
a b
we see that the intercepts which the st. line
Ax + By + C = makes on the axes of x and y are
respectively
c , c
 a and  g.
The same results are obtained by alternately letting
y = and x = in Ace + B^ + C = 0.
32. To obtain the result of § 26 from the general
equation of the first degree.
Let (x v y x ), (x 2 , y 2 ) be fixed points on the st. line
represented by the general equation, and we have
(1) Ax + By + C = 0,
(2) Aa^ + By 1 + C = 0,
(3) A.r 2 + By 2 + C = 0.
From (2) and (3),
(4) _A_ = — B — = 5 .
2/1 — 2/2 as,  a* a^ 2  a^
.*. , from (1) and (4),
as (2/i  2/2) + 2/ O2  aO + x x y. 2  x 2 y x = 0.
This equation is seen to be the same as
x  x, = y  y x
asi  ^2 2/i ~ 2/2'
when the latter is cleared of fractions and simplified.
32
ELEMENTARY ANALYTICAL GEOMETRY
3.3. To find the equation of a st. line in terms of
its inclination to the axis of x and its intercept on
the axis of y.
y
B
N
M
* — ^
v *
Fio. 21.
Let the st. line cut Ox, Oy at A, B respectively,
L BAx = a, and OB = b.
Take any point P (x, y) in the line, and draw PM 1
Ox, PN _L Oy.
m r,™ BN BO  PM h  y
Tan BPN  ^ = om = 5—
But, tan BPN = tan PAM =  tan a.
b  y
tan a
x
and :. y = x tan a + b.
If we let tan a = m, the equation becomes
y = mx + b.
THE STRAIGHT LINE 33
In this equation m is called the slope of the line,
and the Z a, or tan~ l m, is always measured by a
rotation in the positive direction from the positive
direction of Ox, i.e., the Z is traced out by a radius
vector starting from the position Ax and rotating
about A in the positive direction to the position AB.
Note. — For oblique axes the proof and result are different
from those given above for rectangular axes.
(3C) The equation Ax + By + C = may be changed to
A C
y =  b x  b'
from which by comparison with
y = tux + b,
it is seen that the slope of the st. line Ax + By { C = §
_ =» and its intercept on the axis of
is —
B
3   jcj) ^yfmjjL
34
ELEMENTARY ANALYTICAL GEOMETRY
35. To find the equation of a st line in terms of
the j_ on it from the origin and the l made by a
positive rotation from Ox to this _l_.
y
o
N ^S^
X
Fio. 22.
Let the jl OM from O to the line = p, and
I #OM = a.
Take any point P (x, y) in the st. line.
Draw PN j. Ox, NRi OM, MH I! Oy to meet NR at H.
OR + RM = p.
OR = ON cos RON = x cosa.
RM == MH cos RMH = PN cos MOy = y sum.
.'. x cos a + y sin a = p.
u Note. — i^or oblique axes the proof and result are different
from those given above for rectangular axes.
36. To reduce the equation Ax + By + C = to
the form x cos a + y sin a = p, where p is always a
positive quantity.
The equations
x cos a + y sin a — p =
Ax + By + C =0
will be identical if
cos a sin a — p
 f / tit—
*
*B 2
THE STRAIGHT LINE
If C is a positive quantity,
p _cos a _sin a _ y ' cos 2 « + sin 2 «
C _ ~^~A ~^TT ~
t/a 2 4 B 2 y A 2 + B 2
 A  B _ C
.'. cos a = ■ , ' , sin a = — , and p =
y'A* + B 2 i/A 2 + B 2 V A 2 + B 2
If C is a negative quantity, these results should be
written
A B C
cos a ==r , sin a _ , p
 / A 2 I Di ./A2I D 2 ■*•
y A 2 + B 2 ' i / A 2 4 B 2 '^ /A 2 + B 2
Thus the equation is
AX =pBy = _±C
V/A 2 + B 2 v'A 2 + B 2 /A 2 + B 2 '
\
the upper signs being taken when C represents a
positive quantity and the lower signs when c repre
sents a negative quantity.
37. Ex. 1. Reduce the equation Sx + 4y — 12 = to
the form x cos a + y sin a = p.
Here j/j^T"! 2 = l/25 = 5.
Dividing the given equation by 5
3 4 12
cos a — , sin a = , and .*. a = tan ( \ while the
12
X from the origin on the line is —
Ex. 2. Reduce the equation x — y \ 7 = to the
form x cos a + y sin a = p.
t" fr^>< ± "&X
, "~ — » ~i — r~r= — :
* *B X
36
ELEMENTARY ANALYTICAL GEOMETRY
Here /P + 1' = V 2.
Dividing the given equation by — y/2
05 1/
i/2 i/2
7
~i/2
or,
35 COS 135°
+ y
sm 135°
7
~ i/2
i.e
, a =
135°
, and the
J_ from the
origin
line
. 7
is — 
i/2
38. To find the equation of a st. line in terms of
the coordinates of a fixed point on the line and
the z which the line makes with Ox.
Let Q (x v y{) be the fixed point and 6 the Z.
Take any point P (x, y) on the line and let QP = r.
Draw PM, QN j_ Ox and QR ± PM.
qr = pq cos PQR.
QR = NM = x — x u and Z PQR = Z Q.
X — x x
it = r 
cos 6
PR =
PQ
sin
PQR.
PR =
PM
 RM =
PM
2/ 
sin
2/i
e '
= r.
x 
*l
_ y
yi
= ;
THE STRAIGHT LINE 37
QN = y  y v
cos 6 sin
This form will frequently be found useful in pro
blems that involve the distance between two points on
a st. line.
If cos 6 = 1 and sin 6 = m, the equation becomes: —
** i _ yyi _ r
1 m
I and m are called the direction cosines of the st.
line, and I 2 + m 2 = i.
39. For convenience of reference the different forms
of the equation of the st. line are here collected : —
(1) Ax + By + C = 0.
(2) +
a
y
b
= 1.
(3) X "
x x
_ 2/ ~ 2/i
2/i  2/2
+ y sin a
C8) *! 
(4)2/ =
(5) x cot
x 2
nt.r
1 a
= 7?.
(6) " 
If the st. line passes
e convenient form : —
x i  y  2/1 _
m
through the >
= r.
arigin (4) takes
0)y =
in,'
38 ELEMENTARY ANALYTICAL GEOMETRY
If the st. line joins the origin to a fixed point
(Ph> Vi)> we get, by letting x 2 = y 2 = in (3): —
(8) X  = V 
If the st. line passes through (x u y x ) and its slope is
m, the equation is easily seen from (4) to be
(9) yy 1 = rrb{x x x ).
40.— Exercises
1. Name the constants and variables in each of the nine
equations of § 39. Explain the meaning of each constant.
Which of these equations are of the same form for rectangular
and oblique axes 1
; '2. , Draw the following st. lines on squared paper : —
(a) x + 2y = 8 ; (b) 3x  ly =  5 ; (c)  +    lj
(d) 2x + 3y = 13; (e) Sy = 4x.
3. Find the equation of the st. line.
(a) through the origin and making an Z of 30° with Ox;
(b) through the origin and making an Z of 120° with Ox ;
(c) through (0, 5) and making the Z tan _1 f with Ox ;
(d) through (0, 3) and making the / cos 1 £ with Ox;
(e) through (  3,  4) and making the Z. 45° with Ox.
4. In the A of which the vertices are ( — 2, 5), (3,  7),
(4, 2)
(a) find the slope of each side;
(b) show that the medians are concurrent and find the
centroid.
5. O (0, 0), A (6, 0), B (4, 6), C (2, 8) are the vertices
of a Quadrilateral. Show that the st. lines joining the
THE STRAIGHT LINE 39
middle points of OA, BC, of AB, CO and of OB, AC are
concurrent, and find the coordinates of their common point.
•fl^Find the equation to the st. line through (  4, 3) that
cuts off equal intercepts from the axes.
7. Find the length of the _L from the origin to the line
Bx + 7y = 10; find the Z which this ± makes with Ox.
v 8. What is the condition that the st. line Ax f By + C =
may
v (a) pass through the origin;
■■(b) be  Ox ;
J (c) be  Oy ;
 (d) cut off equal intercepts from the axes ;
v (e) make I 45° with Ox;
/ 9T'~'\Vhat must be the value of m if the line y = mx j 7
passes through (  2, 5) 1
y 10.' Find the values of m and 6, if the st. line y = moo + 6
passes through (  2, 3) and (7, 2).
v [11. Find the values of a and b, if the st. line _ f ^ = 1
a b
passes through (  2,  5) and (4,  2).
JL2. Show that the points (4a, 36), (2a, 0), (0, 36) are
in a st. line.
\ 13. Show that the intercept made on the line x = k by
the lines Ax \ By + O = Q and Ax + By f C' = is the
same for all values of k.
14. P (x v y x ) is any point and the lines Ax f By + C =
cuts Ox, Oy at N, R respectively. Show that /\ PNR =
2^B (Ax i + B ^i + C >'
40
ELEMENTARY ANALYTICAL GEOMETRY
The Angle Between Two Straight Lines
41. To find the Z between two st. lines whose
equations are given.
(i) Let the given equations be y = m x x + 6 X and
y = m.jX + b 2 .
Let AB be the line y = m x x + b x and AC be the line
y = m.jX f b 2 when B, C are on the axis of x. Let
L BAC = 6.
Then n\ = tan AB£, m 9 = tan AC#.
.*. tan 6
z 6 = z ABj? — z ACr
fcm AB21 — ton AC.s
tc^Grto^*
_ HC*
w.
= tan
1 + tan AB£ . tan ACx 1 + m 1 m 2
m, — m.,
1 + m^
(fn)) Let the given equations be Ax + By + c = and
A,a; + B,2/ + Cj = 0.
These equations may be changed to
„=**§ and y = £*
THE ANGLE BETWEEN TWO STRAIGHT LINES 41
.*. writing — — for m, and — — 1 for m 9 in the above
result, the Z between the lines
_ ^ + A x
+ BBj
(^42) Condition of Parallelism. If two st. lines are
, they make equal Zs with the axis of x, i.e., their
slopes are the same.
.*. , if their equations are y = m r « + b x and y =
rn<£C f 6 2 , the condition is
m 1 = m 2 .
If their equations are Ax + By + C = and
A x x + B^ + C x = 0, the condition is
_ A _ _ A,
~ B ~ b/
or, AB X  AiB = 0.
A B
This may also be written t = o ', and we see that
the equation ax + by = k can be made to represent
an infinite number of  st. lines by giving different
values to k ; as : — ax + by = k x , ax + by = k 2 , etc.
(J3. Condition of Perpendicularity. If the st. lines
y = m x x f b x , y = m.pc + b 2 are j_,
tow _1 — ! ?_ _ «_^
1 + mjWig 2
1 ■+■ m x m<,
42 ELEMENTARY ANALYTICAL GEOMETRY
This will be true if 1 + m x m 2 = 0, and .. the
required condition is
n^m* =  1.
Similarly, if Ax + By + C = and A x x + B x y + C L =
are J_.
AA X + BB : = 0.
The st. lines
Ax + By + C =
Bx  Ay + Cj=
satisfy the above condition and .*. are ± to each
other.
44.— Exercises
1C Find the L between the st. lines
(a) 2x  3y = 9 and x + 5 y = 11 ;
(6) 3x + 5y = 12 and (17/3 + 30) x + 33y = 19;
■ (c) by = 3x + 12 and 5cc + 3y = 17;
(J) ix + ly = 13 and 3* ?/ = 6.
2. Find the equation of the st. line  to 6a?  ly = 13
and passing through (  2,  5).
Solution: — The required equation is
6 (x + 2)  7 (y + 5) = ;
i.e., 6x  ly = 23.
3. Find the equation of a st. line through (  3,  5)
and  to 9x + 4y = 18.
4. Find the equation of the st. line drawn through (  2,
 5) and J_ to 6a;  ly =13.
Solution: — The required equation is
7 (x + 2) + 6 (y + 5) = ;
i.e., lx + 6y + 44 = 0.
5. Find the equation of the st. line drawn through (4, 2)
and J_ 3x 2y = I.
EXERCISES
43
^6. Find the equation of the st. line passing through
(3, 5) and  to the st. line joining (2, 6) and (7, 1).
^m Find the equation of the st. line passing through (2, 6)
and JL the st. line joining (  3, 5) and (7,  1).
V8f Find the equations of st. lines drawn through (5, 7)
which make Zs 45° and 135° with Ox.
\>9. Find the equations of st. lines drawn through (  5,  3)
which make ^s 30° and 150° with Ox.
v^lCJ. Show that the _l_s from the vertices of the A (1,  3),
(  5,  2), (4, 7) to the opposite sides are concurrent ; and
find the coordinates of the orthocentre.
11. Show that the J_s from the vertices of the A (0, 0),
(a, 0), (b, c) to the opposite sides are concurrent ; and find
the orthocentre.
12. Find the ratio into which the _L from the origin on
the st. line joining (2, 6) and (5, 1) divides the distance
between these points.
03.;Find the equations of the st. lines which pass through
(h, k) and form with y = mx + b an isosceles A of which
the vertex is at the given point and each base Z = a.
n,
Solution : — Let y  k = M (x  h) represent one side of the A
where the value of M is to be found.
44) ELEMENTARY ANALYTICAL GEOMETRY
Tan a ■■
M
m M
m + tan n
1  m tan a
.'. the equation of this side is
7 m + tan a , ■,,
y  h= . (x  h).
1  m tan a
If for a we substitute tQ8°  a, the equation of the other side is
found to be
yk= »  *» * (X  A).
1 + m <om a
^k Find the equations of the st. lines passing through
(2, 8) and making an I of 30° with 3x  I2y = 7.
^l&T Find the equations of the st. lines passing through
(1, 2) and making an Z of 45° with * +  = 1.
16. Show that the equation of the st. line through (a, b)
and making an Z of 60° with x cos a + y sin a = p is
y — b = {x  a) tan (u ± 30°).
17. Show that the right bisectors of the sides of the A
(0, 0), (a, 0), (b, c) are concurrent ; and find their point of
intersection.
18. Find the equation of a st. line J_ to Ax + By f C
= 0, and at a distance p from the origin.
PERPENDICULARS
1,5
Perpendiculars
(45) To find the length of the _L from P (x lt y x ) to
A^ + Bi/ f C = 0.
y
Draw PM _L the given st. line. Join P to N, R
the points where the given st. line cuts Ox, Oy.
A PRN =1PM. RN.
ON = — ^ and OR = 
RN
B 2 .
By the formula of § 17
A
=  2Xi (A ri + B ^ + c >
•'• i PM • ?b\/ A2+ B ' = " 2a¥ (a *> + B ^ + c >
A^ + By, + c
;. PM = 
The length of the _]_ is .'.
ax x + By x + C
•a» 4 B2
/A*+B»
46
ELEMENTARY ANALYTICAL GEOMETRY
In the diagram AB is the line represented by the
equation Ax + 5y — 20 = 0.
, .
1
si; y 
%
>«h ^ s
*V
5^>i
N
v
N ^
N>
..
S
Tj
V
 ^ ,
V 5 J
"v
If
 20.
 25.
 27.
+ 30.
+ 35.
Fio. 27. (Unit = T '„ inch.)
the expression Ax + 5y  20 we substitute
the coordinates of points O, P, Q, R, S which are not
in the line, the following results are obtained.
For o (0, 0), Ax + 5y  20 =
 P (5, 5), 4x + 5y  20 =
m Q (8, 5), 4cc + 5y  20 =
.. R (5, 6), Ax + 5y  20 =
ii S (10, 3), Ax + 5y  20 =
In these results it will be observed that: —
For the origin the sign of the value of the expression
is the same as the sign of the absolute term.
For other points that lie on the same side of the
given st. line as the origin the signs of the values of
the expression are the same as the sign of the result
for the origin; while, for points on the side remote
from the origin the signs of the values of the expression .
are different from the sign of the result for the origin.
PERPENDICULARS
47
A formal proof of these properties is given in the
next article.
46. To prove that the sign of the expression A*
f By + c is different for points on opposite sides
of the line ax + By + G = 0.
y
3
Jk"^
N
Q
^ f
\
1
R
— ?
p (^h' Vi)> Q ( x 2> Vd are an y points on opposite sides
of Ax + By + C = 0.
Draw PM, QR _L Ox and let them cut the given line
at N, S.
Vi
PM
PN + NM ; y, = QR = SR  SQ.
.. AX X + By x + C = AiCi + B.PN + B.NM + C,
and Ax 2 + By 2 + C = Ax. 2 + B.SR  BSQ + C.
But, V N and S are both on the given line,
Ax x + B.NM + C = 0,
and Ax, + B.SR + C = 0.
.*. A£Ci + By x + C = B.PN,
and Ax 2 + By 2 + C = — B.SQ.
.*. , since PN, SQ are both taken as positive quanti
ties, Ax x + By x f C and Ax 2 + By 2 + C have opposite
signs.
48
ELEMENTARY ANALYTICAL GEOMETRY
When x = and y = the expression Ax + By 4 C
becomes C, .". a point whose coordinates when substi
tuted in Ax + By + C gives the same sign as C
is on the same side of the st. line Ax + By + C =
as the origin.
47. Sign of the Perpendicular. It follows from
the preceding article that, if the positive sign is
always taken for j/A" 2 + B, when the sign of
A£C, + B//, + C
/A2 4 B2
is the same as the sign of C, the point (x v y x ) and
the origin are on the same side of the line Ax + By
+ C = 0; and when the sign of this fraction is
different from that of C, the point (x v y x ) and the
origin are on opposite sides of Ac + By + C = 0.
o find the equations of the bisectors of the
zs between the lines Ax 4 b</ 4 c = and A x jr 4
Btf 4 C x = 0.
PERPENDICULARS 49
The Is to the st. lines from any point P (x, y) on
either bisector are equal to each other,
.*. the required equations are
Ax + By + c = ^ Aj x+_B 1 y_+_c 1
^A 2 + B 2 V A x 2 + Bf
If the equations are so written that C and C x have
the same sign and P, is on the bisector of the L that
contains the origin, the JLs from P, have the same
sign as the J_s from the origin on the lines, and the
equation of the bisector is
Ax + By + C A x x + B l2 / + C a
V A 2 + B 2 VAs + Bj 2
If P is on the bisector of the i_ which does not
contain the origin, the Is from P have opposite signs
and the equation of the bisector is
Ax + By + C A X X + B x y + C x
l^A 2 + B 2
50
ELEMENTARY ANALYTICAL GEOMETRY
(49.^ To find the distance from (a, b) to Ax + By + c =
in the direction whose direction cosines are /, m.
The equation of the st. line passing through (a, b)
in the given direction is, by § 38,
x — a y — b
I 711
:. x = a + lr, y = b + mr.
Substituting these values for x and y in Ax \ By +
C = 0,
A<x + Air + B6 + Bmr + C = 0.
_ Aa + Bb + c
Al + Bm
50. The length of the J_ from (a, b) to Ax + By +
C — may be deduced from the result of § 49.
a y  b
For, if
and Ax + By + C = are
_]_ to each other,
A _ B
I m
since I 2 + m 2 = 1.
A£ + Bm
P + m 2
Also each of these fractions
= Al + Bm,
VA 2 + B 2
l/A 5
B 2 .
VI 2 + m l
:. Al + Bm = VA 1 + B 2 ,
and the length of the ± is
Aa + Bb + C
i/A 2 + B 2
PERPENDICULARS 51
51. To find the equation of a line passing through
the intersection of two loci.
The equation
k (ax + By + C) + I (A x « + B lV + CO = 0, (1)
being of the first degree in x and y represents a st. line.
If (x v 2/ x ) is the point of intersection of
Ax + By + C = (2)
and A^ + B x y + C 1 = 0, (3)
the values x v y x substituted for x, y will plainly
satisfy equation (1), and .*. the st. line (1) must
pass through the point of intersection of the st. lines
(2) and (3).
From the same reasoning the following more general
theorem is seen to be true: —
If two equations are multiplied by any numbers
and the results either added or subtracted, the re
sulting equation represents a locus that passes
through the point (or points) of intersection of the
loci represented by the first two.
52. Example — Find the equation of the st. line passing through
the intersections of 17x  ly = 9, 3x + 19y = 34 and J. to llx 
4y = 13.
17x  ly  9 + I (3x + 19y  34) =
is a st. line passing through the intersection of the first two lines.
This equation may be written
(31 + 17) x + (19*  7) y  341  9 = 0.
If this line is _1_ to llx  4y = 13,
11 (3/+ 17)  4 (19/  7) =
.. 1 = 5,
and the required equation is found to be
32x + 88y = 179.
52 ELEMENTARY ANALYTICAL GEOMETRY
53. To find the condition that the three st. lines
a x x + b x y + c x = (1)
a 2 x + b 2 y + c 2 = (2)
a s x + b 3 y + c, = (3)
may be concurrent.
If the three st. lines are concurrent, the coordinates
of the common point satisfy the three equations.
For that point, from (1) and (2),
s = y = 1
b x c 2 — b 2 c x c x a 2 — c 2 a x a x b 2 — a 2 b x
Dividing the terms of (3) respectively by these
equal fractions,
a 3 (6 X c, — b 2 c x )  & 3 (c x a 2 — c 2 a x ) + c. d (a x b.> — a 2 6 X ) = 0.
This is the relationship that must hold among the
constants in order that the lines may be concurrent.
EXERCISES 53
54. Exercises
?{. Find the length of the __
(a) from (  4, 7) to 5x  1y = 4 j
(ft) from ( 4, 3) to +  = 1 ;
(c) from (3,  2) to y = 7x + 1 ;
"(c?) from (  2,  7) to the st. line joining (5, 3) and
(3, 7);
"(e) from the origin to the st. line joining (7, 0) and
(0, 5).
/2. Find the distance between the [ lines ix  3y = 9,
Ax Zy = 2.
x 3. Find the distance between the  lines ax f by + c x
^4. Find the point in the line — f ^ =  1 such that its
J_ distance from the st. line joining (2, 7), (5, 3) is 8.
* 5. Find the equation of the st. line through the intersec
tion of 3x  2y = 12, 5x + iy = 9 and  to  + V  = 1.
(See §§51 and 52).
^6". Find the equation of the st. line joining the origin to
the intersection of — f f = 1 an( l T +  = 1.
a b o a
lZ?r Find the distance from the point of intersection of
7x  by = 13, 4ic + 9y = 43 to the line 12a; = by.
J^T'Find the equation of the st line passing through the
intersection of y = mx \ c, ?/ = n^a; + c x and J_ to
a; y
7, + f = L
54 ELEMENTARY ANALYTICAL GEOMETRY
ytf. Show that the st. lines x J 2y = 5, 2x + 3y = 8,
3x f j/ ■=> 5, x + y = 3 and 2a;  y = are concurrent.
10. Find the condition that the lines ax \ hy \ g = 0,
hx {■ by \ / = 0, ya: + /y f c = are concurrent.
lif Find the equation of the st. line passing through the
intersection of y = mx \ c, y = m^x + C\ an d also through
(a, b).
\12. Find the equation of the st. line joining the origin
to the point of intersection of Ax f By + C = and
* x x + B x y + Cj = 0.
"13. Find the distance from the orthocentre of the A
O (0, 0), A (8, 0), B (3, 5) to the st. line AB.
14. Plot the lines 2x  3y = 1, 3a; f y = 7 on squared
paper and find the intercepts that the bisectors of the Ls
between them make on the axis of y.
"' 15. Find the equations of the bisectors of the ^s between
5x  12y = 17 and 8a; + 15y = 31.
16. Show that the bisectors of the supplementary Zs
between y = mx J « and y = m r x f a x are __ to each
other.
17. Find the equations of the bisectors of the L& between
the st. lines joining (4, 5) and (  5, 2) respectively to
(3, 7).
v/18.' Show that the st. lines x f 6y = 15, 2x  5y f 4 =
and 9 a; + y = 29 are concurrent.
19. The sides of a A are 3x + 4y = 15, 12a;  5y = 17,
24a; j 7y = 30. Plot the lines on squared paper and find
the point where the bisectors of the interior Zs of the A
intersect.
EXERCISES 55
,2©T Find the equation of the st. line passing through the
intersection of the lines 2x  3y — 6 and 3x + 4y = 18,
and also through the middle point of the st. line joining
(1, 2) and (3, 4).
i2?. Find the distance from (5, 3) in the direction in
which the slope is — = to the line Ix  lly = 13.
V3 J
j^T Find the distance from (4, 6) in the direction of
which the slope is 1 to the line 1_ V = \
r 2^3
Draw the diagram on squared paper.
25T The sum of the distances from a point to the lines
x + 2y = 7, 5x  2y = 11 is 7. Show that the locus of the
point is a st. line which makes equal Zs with the given st.
lines.
^JteT Find the distance between the  lines _ i £L _ c
a o
25. Find the equation of the st. line passing through P
(2, 5) and cutting Ox at A, Oy at B so that AP:PB =7:3.
26. Find the equations of the st. lines passing through
(4, 7) and making an Z of 45° with 3x  10 y = 8.
27. Find the equations of the st lines passing through
(  4,  7) and forming an equilateral A with 3x  2y = 7.
28. Find the equations of the st. lines drawn  to
5x  I2y = 9 and at a distance 5 from it.
29 Find the equations of the two st. lines which pass
through (4, 7) and are equally distant from A (7, 3),
B (3,  1). Find also the distances from A and B to
these lines.
56 ELEMENTARY ANALYTICAL GEOMETRY
30. Find the point in 4x  3y = 12 which is equally
distant from (2, 7) and (4,  1).
31. Having given the length of the base and the difference
of the squares of the other two sides of a A, prove that the
locus of its vertex is a st. line _L the base.
32. Find the equations of the st. lines which are at a
distance 2 from the origin and which pass through the
intersection of a; 7yf 11 =0 and 3a? + 4y  17 = 0.
33. Find the equation of the st. line  to Ax + By f
C = and at a distance p from the origin.
34. Find the equation of the st. line passing through
(h, k) and _L to Ax + By f C = 0.
35. The equations of the sides of a A are 5x { 3y 
15 = 0, 2x  y + 4 = 0, Sx  1y  21 = 0. (a) Show
that the _l_s from the vertices to the opposite sides are
concurrent and find the coordinates of the orthocentre. (b)
Show that the right bisectors of the sides are concurrent
and find the coordinates of the circumcentre. (c) Show
that the centroid is at a point of trisection of the st. line
joining the orthocentre to the circumcentre.
CHAPTER m
The Straight Line Continued. Transformation
of Coordinates
55. An equation of the second degree may represent
two st. lines.
For example, 2x 2 — 5xy + Sy 2 = is the same as
(x —y) (2x — 3y) = 0, and will be true for all values
of x and y which make either of the factors x — y or
2x —3y equal to zero, and .". all points on the st.
lines x — y = 0, 2x — Sy = are on the locus represented
by 2x 2  hxy + 3y 2 = 0.
Similarly, an equation of the third degree may
represent three st. lines, one of the fourth degree may
represent four st. lines, etc.
56. The general equation ax 2 + 2hxy + by 2 =
represents two st. lines passing; through the origin.
Solving as a quadratic in x
h ± Vh  ab
— y>
a
from which it is seen that the given equation is
equivalent to
{ax + hy + y Vh 2 — ab} {ax + hy — y Vh 2 —ab) = 0,
and .'. represents the two st. lines
ax + hy + y Vh 2 — ab =
ax + hy — y Vh 2 — ab = 0,
both ot which pass through the origin.
57
58 ELEMENTARY ANALYTICAL GEOMETRY
If h 2 > ab, the lines are both real.
If h 2 = ab, the lines are coincident.
If h 2 < ab, the lines are imaginary, and we have
two imaginary sfc. lines passing through the real point,
(0, 0).
57. To find the z between the two st. lines
represented by ax 2 + 2hxy + by 2 = 0.
The given equation may be written
2/H2^y + * x 2 = 0.
If y — m x x and y — m^c are the factors of the
expression on the left hand side of this equation.
2h a
m x + ra 2 = j , m^z = ^.
.*. mj 2 j 2 m^g + m 2 2 =
4A 2
b 2 '
4 ra^j =
.*. K  ™ 2 ) 2 = 62
b '
ab)
2 i//t 2 
and m, — m, = ; —
then 6 is the Z between the st.
ab
lines, by §41.
m, — m 2 2 Wi 2 —
tort = — 3 L. =
1 + m[m 2 o
ab b
X a + b
2 Vh 2  ab
a + b
THE STRAIGHT LINE CONTINUED 59
...fltoa 8 ^'^
a + b
Condition of perpendicularity. If = 90°,
tan 6 = oo. This will be the case if
a + b = 0.
58. To find the equation of the St. lines which
bisect the zs between the st. lines represented by
ax 2 + 2 hxy + by 2 = 0.
Let the given equation represent the st. lines
y — m r £ = 0, y — m 2 x = 0, so that
2h a
m x + m., = — __, m^m^ = r .
o b
The equations of the bisectors of the Zs between
these lines are
V  m i x , V ~ ™ >& y  m x x y  m,x
z H , = and , — =0.
vi+mf y\+m 2 2 ^l+mf Vi+ m2 2
These equations may be combined into
(y  m x xf _ (y  m 2 x) 2 =
1 + m 2 1 + w 2 2
Simplifying and dividing by ra 2 — m v
(m x + w^) y 2  2 (mjm 2  1) xy — (m 1 + m 2 ) x 2 = 0.
Substituting and multiplying by b.
h (x 2  y 2 )  (a  b) xy = 0.
60 ELEMENTARY ANALYTICAL GEOMETRY
59. To find the relationship that must connect
the constants in the equation
ax" + 2 hxy + by + 2 gx + 2fy + c =0
in order that this equation may represent two st.
lines.
If the given equation represents two st. lines, it
must be equivalent to two equations of the form
y — m^x — b x = 0, y — m 2 x — b 2 = 9, from either of
which y can be expressed in terms of the first degree
of x.
Solving the given equation for y, we obtain
 (hx + f) ± V(hx + fY  b (ax 2 + 2 gx + c)
y =  5 —
In order that these values of y may be in terms of
the first degree of x, the expression under the radical
sign must be a perfect square ; i.e.,
(/i 2  06) x 1 + 2 (hf  bg) x+f be
is a perfect square for all values of x.
.. (hf  bgf = (h*  ab) (f  be).
Simplifying, we get the condition in the form
2 fgh  ap  bg  eh* + abc = 0.
60. — Exercises
1. Show that the following equations represent two st.
lines and find the separate equations of the lines : —
(a) x 2  (a f b) x =  ab ; (b) x 2  y 2 = ;
(c) x 2  Zxy = ; (cl) 8x 2 f 3y 2 = 1 Oxy ;
(e) xy + bx = ay + ab ; (/) 3x 2  lOxy + 3y 2  llx
 7y  20 = 0.
EXERCISES 61
2. Show that 2x 2  Ixy f 6y 2 + 2s  5y  4 = repre
sents two st. lines and find the slope of each.
3. Interpret the locus represented by xy = 0.
4. Find the Zs between the st. lines in 1. (d), (e) and (/).
5. Find the condition that axy + bx f cy + d = may
represent two st. lines.
6. Find the value of B for which the equation 3x 2 
lOxy + By 2  1x  2y = 21 will represent two st. lines.
7. Find the single equation which represents the two st.
lines passing through (5, 3) and making an equilateral A
with the axis of x.
8. Prove that y 2  Ixy sec a f x = represents two st.
lines through the origin and inclined to each other at an
L = a. Show also that one of these lines makes the same
Z with the axis of x that the other makes with the axis
of y
ELEMENTARY ANALYTICAL GEOMETRY
Transformation of Coordinates
61. It is often necessary to change the coordinates
involved in a problem into a different set which are
referred to axes drawn
(a) from a new origin, or
(b) in directions different from the original axes.
62: To change from a pair of axes to another
pair which are  to the former, but have a different
origin.
V
Y
ij<)
N
Q
X
R
r
^ f
Fig. 30.
Q (h, /,•) is the new origin.
Let P (x, y) be any point referred to Ox and Oy
and X, Y the coordinates of the same point referred
to the new axes QX and QY.
Draw PNM ± to Ox and QX, and let YQ cut Ox
at R.
X = OM = OR + QN = h + X.
y = PM = QR + PN = k + Y.
TRANSFORMATION OF COORDINATES
63
Thus, if for x, y respectively we substitute h + X,
k + Y in any equation the origin is changed to the
point (h, k).
To return to the original origin the substitutions
would be X = x — h, Y = y — k.
(^63^To change the direction of the axes, without
changing the origin, the axes being rectangular.
y
^^^.CC
s
I
i
i
^y
y/O
I
A R
X
Let P (x, y) be any point referred to Ox, Oy ; and
X, Y the coordinates of the same point referred to
axes OX, OY such that I XOa; = a.
Draw PM J_ Ox, PN J_ OX, NR __ Ox, NS _L PM.
Z NPS = 90°  Z NAP = 90*  Z MAO = a.
x = OM = OR  NS = X cos a  Y sin a. j
V = PM = NR + PS = X sin a + Y COS a. I
T*
Thus, if for x, y we substitute respectively X cos' a
— Y sin a, X sin a + Y cos a, the axes are rotated in
the positive direction through an Z a .
64 ELEMENTARY ANALYTICAL GEOMETRY
64. By § 35, in an equation of the form x cos a + y
sin a = p, the length of the J_ from the origin on the
st. line is the absolute term p.
If, without changing the direction of the axes, the
origin be transferred to (x v y^) the equation becomes
(x + x{) cos a + (y + 2/1) sin a = p,
i.e., x cos a + y sin « = p — x 1 cos a — y x sin a.
The new equation is of the same form as the old
one except that the absolute term is now p — x 1 cos a
— y x sin a.
This absolute term is then the length of the _l_
from the new origin to the st. line; or, reverting to
the original origin, the length of the JL from (x v y x )
to the line x cos a + y sin a = p is p — x x cos a — y l
sin a.
This is the same as the result that would be
obtained by using the formula of § 45.
65.— Exercises
<f! "What does the equation 2a: 2  llxy f 12y 2 + 7x 
13y f 3 = become when the origin is changed to the
point (1, 1) the directions of the axes being unchanged?
^J2. Transform the equation x 2 f xy  7x  iy + 12 =
to  axes through (4,  1).
3. Find the point that must be taken as origin, the
directions of the axes being unchanged, in order that the
terms of the first degree in x and y may vanish from the
equation x 2 + y 2 + 5x  9y + 17 = 0. Find also what the
equation becomes.
EXERCISES 65
4. Show that the terms of the first degree in x and y
will vanish from the expression ax 2 + 2hxy + by + 2gx +
/hf ~ bg hg  af\
2fy + c, if the origin be changed to (^ — , 7 — J,
the directions of the axes being unchanged.
^6. Transform the equation Ax + By + C = by rotating
the axes through an L of 30°.
6. Find what the equation x 2  y 1 = a 2 becomes when
the axes are turned through an L of 45°, the origin
remaining the same.
/^. Show that the equation x 2 + y = a 2 is not changed
when the axes are turned through any L a, the origin
remaining the same.
^8. Find what the equation 33a; 2  34 V^xy  y 2 =
becomes when the axes are turned through an L of 60°,
the origin remaining the same.
9. Find the smallest positive L through which the axes
must be turned in order that the coefficient of xy in the
equation 59x 2 + 24 xy + 66y' 2 = 250 may vanish ; and also
find what the equation becomes.
10. Show that the term involving xy in the expression
ax 2 \ 2 hxy \ by 2 will vanish, if the axes are turned
through the A
2 a  b
66 ELEMENTARY ANALYTICAL GEOMETRY
6G.— Review Exercises
1. Find the distances between the following pairs of
points : —
(a) (2, 7), (6, 2);
(6) (2a + 6, a  26), (a  b, 3a + b) ;
(c) (a cos a, a sin a), (b cos a,  b si/i «).
Verify the result in (6), on squared paper, when a = 1,
b =  2.
^2; A ( — 5, 1), B (4, 6) are two given points, P is
taken in AB and Q in AB produced such that AP : PB =
AQ : QB = 5:3. Find the coordinates of P and Q.
V J$". Find the area of the A of which the vertices are
(3a, 2b), (2a, 36) and (a, 6).
"">£ Find the area of the A contained by the lines
2x + Uy f 43 = 0, 9a: + 8y  14 = and 7x  ty +
2G = 0.
jf. Find the _L distance from (  2, 3) to the line
3 2
Should the result be considered positive or negative and
why 1 ?
6. Find the condition that the three points (x v y 1 ),
( x 2' yi)' ( ,r 3> vi) ma y ^ e * n a s ^ ^ me 
Jfc Find the locus of a point such that the square of its
distance from ( — 3,  7) exceeds the square of its distance
from (5, 0) by 43.
8. Prove that the equation Ax + By + C = represents
a st. line.
9. Find the equation of the st. line which is equidistant
from the  lines ax f by = c, ax f by = d.
REVIEW EXERCISES 67
10. Find the equation of a st. line which makes an Z a
with Oy and cuts off an intercept b from Ox.
•^Vt Show that the st. lines Ax f By + C = 0, A,x + B^
■f C, = are , if AB X = AiB.
1*2. Explain the meaning of the constants in the equations
x  h y  k
cos sin 6
^\&. Show that the line y = x tan a passes through the
point (a cos a, a sin a), and find the equation of the _L to
the line at that point.
^]A. Find the Z between the st. line joining (  4, 5),
(5, 1) and the st. line joining (3, 7), (  6,  3).
^lX Find the values of in and b such that the line
y = rax j b will pass through (3,  2) and ( — 1,  5).
\/y!}. Find the equation of the st. line which passes
through (2,  2), and makes an L of 150° with Ox.
17. Find the length of the st. line drawn from (h, k), in
the direction inclined at L a to Ox, and terminated in
the line y = rax f b.
^18. Find the equation of the st. line through (h, k), and
x v
it. + fi.
^>9. Show that the st. lines Aa; + By + C = 0, A^ +
B x y + C x = are _L to eacli other, if AA : f BBj = 0.
y^Q. Write the equation of the st. line which is ± ax 
by — c, and cuts off an intercept = d from Oy.
J%1. Find which of the following points are on the origin
side of * .1 = 1:— (5,3),(2, 8), (2, 2),(6, 14),
3 5
(7, 1"). Illustrate by a diagram on squared paper.
68 ELEMENTARY ANALYTICAL GEOMETRY
22. Show that the points (2, 6), (1, 11), (4, 7), ( 3, 3)
are in the four different angular spaces made by the lines
+ JlMldB£l.
3 T 5 = l and 5  9
Illustrate by a diagram on squared paper.
23. Find the values of a for which the lines 2x  ay 4 1
= 0, rt.r  6?/  1 = 0, 18a; — ay — 7 = are concurrent;
and find also the coordinates of the respective points of
intersection.
24. Show that the condition that the lines ax + by = 1,
cx\dy = 1, hx \ ky = 1 are concurrent is the same as
the condition that the the points (a, b), (c, d), (A, k) are
collinear.
25. Find the L contained by the lines 4x  7y f a = 0,
3a; 4 ii y 4fc = 0.
26. Find the equation of the st. line passing through the
intersection of Ax  ly \ a = and 3x + l\y ■+ 6 =
and making an L of 45 3 with the axis of x.
27. Find the equation of the st. line passing through the
y
b
intersection oi  \ £ = 1, y = mx { c and also through
(d, 0).
28. Show that the equation of the st. line joining the
intersection of x cos a \ y sin a = p, x cos /? + y sin {3 = p
to the origin is y = x tan .
29. Find the length of the _]_ from (a, b) to  f  = 1.
30. Find the equation of the st. line through (  5, 1)
and  to 3.x + Uy = 17.
31. Find the equation of the st. line through (8, —2)
and X to 7x = y f 4.
REVIEW EXERCISES 69
32. Find the coordinates of the four points each of which
is equally distant from the three lines j  _■ = 1,
12 **" 5 " l ' 24 7 = •
'"'33: Find the coordinates of the foot of the _L from
(3, 5) to the st. line joining (1, 2) and (8, 1).
34. Find the separate equations of the st. lines represented
by 3cc 2 + Uxy 2y* = 0.
35. Find the product of the J_s drawn from (3, 2) to
the st. lines represented by ox 2 f Vlxy f 2y 2 = 0.
36. Show that the L between the lines y — ynx J a,
.  . . . m  n
y = nx \ o is tan l .
1 + mn
37. Find the tangent of the L between the st. lines
represented by 5a: 2  &xy y 1 = 0.
t38T Find the equations of the st. lines which pass through
(3, 6), and are inclined at an L of 45° to — + ^ = 1.
337 Show that the st. line joining the point (1, 1) to the
intersection of f r = 1 with  — ( — = 1 passes through
a b b a
the
origin.
40. Show that the equation of the st. line passing through
the intersection of x cos a f y sin a = >p,x cos /3 { y sin ft = q,
and  to x \ y = k is
(x \ y) sin (a  /S) f p (si?i /3  cos /5) f q {cos a  sin a) = 0.
^■tfi. Find the equation of the st. line passing through the
intersection of 5x — ly = 16, 2x  3y = 7 and J_ to
6a; 4y = 19.
70 ELEMENTARY ANALYTICAL GEOMETRY
42. Find the equations of the st. lines drawn through
the vertices and  to the opposite sides of the A of which
the equations of the sides are 3x + Hy = 23, 4x — 9y = 11,
7x 2y = 31.
43. Show that the lines 5.x + y = 4, 2x + y = 2, 3a; + 3?/
= 4 are concurrent ; and find the coordinates of their
common point.
44. Find the equation of the st. line passing through
the intersection of ax + by \ c = 0, fx 4 gy J h = 0, and
(a) also through the origin ; (b) _L to x f y — k.
s4f>. Find the equations of the st. lines which bisect the
Zs between the lines I2x  5y = 17, 8a; + 15y = 13.
46. Find the equation of the st. line which passes through
the point of intersection of the lines 5x  y = 4, 4x  9y
= 11, and is A. to the former.
47. Show that the points (4, 2), (6, 2), (5, 2+^3) are
the vertices of an equilateral A.
48. Find the locus of a point which moves so that the
sum of its distances from the axes is 10. Trace the locus
on squared paper.
49. Find the locus of a point which moves so that the
difference of its distances from the axes is 10. Trace the
locus on squared paper.
50. Find the equations of the st. lines each of which
passes through (  5,  3) and is such that the part of
it between the axes is divided at the given point in the
ratio 7 : 3.
M. Find the equation of the st. line which passes through
(3,  2), and is _L to 4* + y f 12 = 0.
52. Find the equation of the right bisector of the st.
line joining (a, b) and (h, k).
REVIEW EXERCISES 71
J&. Two st. lines are drawn through (0,  3) such tliat
the J_s on them from (  6,  6) are each of length 3.
Find the equation of the st. line joining the feet of the ±s.
54. Find the Z of inclination of the lines ax\by = c,
(a \ b) x  (a  b) y = d.
^J>5. A st. line is drawn through (2,  4) and ± to 7x
— 3y — 11. Find the equations of the bisectors uf tne Zs
between the _L and the given st. line.
56. Find the equation of the st. lines which bisect the
Zs between the lines represented by x 2 f 2xy sec 6 + y 2 — 0.
57. Find the value of h for which the equation 3x 2 4 hxy
— 10y 2 f x + 29y 10 = will represent two st. lines.
58. Show that, if the axes are rotated through an Z of
45°, the terra containing xy vanishes from the equation
x 2 j 2xy sec d + y 2 = 0; and the separate equations of the
a
two st. lines become x = ± y tan —
&9. Three vertices of a gm are (3, 4), (3, 1), (5, 2).
Find the coordinates of the fourth vertex.
60. Prove that the two st. lines which join the middle
points of the opposite sides of any quadrilateral mutually
bisect each other.
61. What must be the value of m, if the line y = mx
5 passes through the intersection of 7x  lly = 14 and
5*4 2y = 11.
62. Find the area of the A contained by the lines x f y
= 12, 2x y = 12, x 2y = 12.
CHAPTER IV
The Circle
67. A circle is the locus of the points that lie at a
fixed distance from a fixed point.
The fixed point is the centre and the fixed distance
is the radius of the circle.
68. To find the equation of a circle having its
centre at the origin.
Let P (x, y) be any point on the circle of which
the centre is O. Let the radius = a.
Draw PM ± Ox. Join PO.
*.' OPM is a rt.zd A,
.'. OM 2 + PM 2 = OP».
x 2 + y 2 = a 2 .
72
THE CIRCLE
73
This being the relation which holds between the
coordinates of any point on the circle and the given
radius is the required equation.
69. To find the equation of a circle, the centre
being at any fixed point (h, k) and the radius equal
to a.
y
""^p^i)
/ u/ ; \
( ' ' «
I M)c
* ,'l
O
h
j M
r
C (h, k) is the centre ; and P (x, y) is any point on
the circle.
Draw PM, CN j. Ox, CL j_ PM. Join CP.
CL = NM = OM — ON = x — h;
PL = PM  LM = PM  CN = y — k.
7 CPL is a rt.^d A,
CL 2 + PL 2 = CP 2 .
.. (x  h) 2 + (y  k) 2 = a 2 .
This is the required equation.
74 ELEMENTARY ANALYTICAL GEOMETRY
7v. If we expand the equation found in § 69, we
obtain : —
a? + 2/2 _ 2Jwc  27cy + h? + k 2  a 2 = 0.
Comparing this result with the general equation of
the second degree : —
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0,
we see that the conditions that the latter should
represent a circle are that the coefficients of x 2 and y 2
should be equal and that the coefficient of xy should
be zero.
Thus the equation
ax 2 + ay 2 + 2gx + 2fy ■+• c =
may be changed to
V V _i_ /„ j_ /V _ (f +/ 2 ac
(• + tr + G' + #
a
from which, by comparison with the formula of § 69,
we see that it represents a circle having its centre at
the point (——, — —) and its radius = —
r V a a' a
71. The general equation of the circle to rectangular
axes is commonly written : —
x 2 + y 2 + 2gx + 2fy + c = 0.
When the circle passes through the origin and its
centre is on the axis of x, the equation of § 69
becomes
(x  a) 2 + y 2 = a 2 ,
or, x 2 + y 4 = 2ax.
EXERCISES 75
72.— Exercises
1. Write the equation of the circle with centre (0, 0)
and radius = i/5.
w 2. Write the equation of the circle with centre (6, 2)
and radius = 3.
v 3. Write the equation of the circle with centre (  5,  1) and
radius = /26. Show that this circle passes through the origin.
4. Write the equation of the circle with centre (  a,  b)
and radius = c. Find the condition that this circle passes
through the origin.
/&. Find the coordinates of the centre and the radii of
the following circles : —
(a), x 2 + y 2  Qx  2y = 15 ; (b), ix f 4y 2 4 7x + 5y
= 16 ;(c), x 2 + y 2 = lix;(d), x 2 + y 2 + 2 by = c 2 .
6. Draw, on squared paper, the circles of which the
equations are : —
(a), x* + y2 = 9 . ( b)j X 2 + y 2 = 8x ; (c), as* + y* + 6y = 7.
7. Find the centre and radius of the circle which passes
through the origin and cuts off intercepts = a and b from
Ox and Oy respectively.
Solution. — Since the circle passes through the origin its equation
must be satisfied by x = 0, y = 0, and .". the absolute term must be
zero. Thus the equation may be written
x 2 + y 1 + 2 gx + 2/y = 0.
Substituting in this equation the coordinates of the points (a, 0),
(0, b) the two equations
a 2 + 2 ga =
6 2 + 2 /& =
are obtained from which a =   , f =  — .
" 2 2
.*. the equation of the circle is
x 2 + y 2 — ax — by = 0,
.*(.i)' t (,iy£±JK_
.". the centre is ( " — ) and the radius = •
V.2 2/ 2
76 ELEMENTARY ANALYTICAL GEOMETRY
' 8. Find the equation of the circle which passes through
the origin and also through (4, 3) and (  2, G).
• 9. Find the equation of the circle which has its centra
on the axis of x and which passes through the points
(5, 3) and (3, 1).
10. Show from the general equation of §71 that three
conditions are necessary and sufficient to determine a circle.
11. Find the condition that the circle x 2 + y 1 + %ff x +
%fy \ c = may have its centre («) on the axis of x ;
(b) on the axis of y.
v 12. Find the equation of the circle which passes through
(3, 1) and (5,  3) and has its centre on the line x  y = 4.
13. Find the equation of the circle having the st. line
joining (7,  5) and (  3,  1) as a diameter.
14. A (a, 0) is a fixed point and P (x, y) is a variable
point such that PO : PA = p'.q. Show that the locus of P
is a circle having its centre on Ox, and dividing OA inter
nally and externally in the ratio p : q.
J )£>. Find the equation of the circumcircle of the A whose
vertices are (3, 4), (  2, 3), (  5,  7).
^>. Find the length of the chord of the circle x 2 + y 2 =
25 cut off by the line 3x + y = 15.
^ Vt. Find the length of the chord of the circle x 2 + y 2 
6x f 14y = 42 cut off by the line x  y = 8.
18. Show that the locus of the centres of all circles which
pass through two given points (p, q), (r, s) is the right bi
sector of the st. line joining the given points.
19. Through the given point P (h, k) a st. line is drawn
cutting the circle x 2 + J/ 2 + 2<?a? f 2fy + c = at A and EL
Prove that PA . PB is constant for all directions of the st.
line.
EXERCISES 77
x  h y  k
Solution : — Take tt = —  — 3 = r as the equation of the st. line.
cos a sui a ^
Then
x = h + r cos 6, y = k + r sin 6.
Substituting these values in the equation of the circle, and simplifying
r 2 + 2 {{h + g) cos d + (k + f) sin d} r
+ h* + k 2 + 2gh + 2fk + c  0.
The value of PA . PB = the product of the two values of r in
this equation
= h 2 + ¥ + 2gh + 2/k + c,
an expression which does not contain and which is .'. independent
of the direction of the line.
20. Find the equation of that chord of the circle x f
y 2 + %9 X + %fy + c = which is bisected at the point (h, k).
Solution:— (First Method). As in Ex. 19, if we take for the
as  h y  k
equation of the chord tt — — : — 7T — r, we get
^ cos sin i) ' 6
r 2 + 2 {(A + g) cos 6 + (k + f) sin 6 } r
+ h* + k 2 + 2gh + 2/k + c = 0.
If the chord is bisected at (h, k) the two values of r got from this
equation are equal in value but opposite in sign, and
.. {h + g) cos 6 + (k + f) sin 6 = 0.
Multiplying the terms of this equation by the equal fractions
, • , the required equation is found to be
cos sin 6 u H
(h + g) (x  h) + (k +f)(y k) = 0.
(Second Method). Let the equation of the chord be
y — k = m (x  h).
The centre of the circle is (  g, /), and the equation of the J.
from the centre to the chord u
m (y + f) + x + g = 0.
The _1_ from the centre bisects the chord, and, ,\ passes through
(h, k).
:. m{k+f) + h + g = 0.
_. h + 9
k +/'
.". the required equation is
(k + g) (x  h) + (k + f) (y  k) = 0.
78 ELEMENTARY ANALYTICAL GEOMETRY
v 31. Find the equation of the chord of the circle x 2 + y 2 
6x  8y — 24 which passes through (5,  1) and is bisected
at that point.
22. From the point P (  3,  7) a st. line is drawn to
cut the circle x 2 + y 2  ix  \0y = 17 at A and B.
Find the area of the rectangle PA . PB.
23. Find the equation of the common chord of the
circles
x 2 + y 2  Sx  6y = 39,
x 2 + y 2 + 6x + % = 56.
24. Find the condition that the common chord of the
circles
x 2 + y i + 2gx + 2/y + c =0,
x 2 + y 2 + 2g'x + 2/'y + c =
passes through the origin.
25. Find the equation of the circle which passes through
the origin and also through (A, k) and (k, h).
TANGENTS
79
Tangents
73. Let APQ be a secant cutting a curve at P and Q.
If the secant rotate about the point P until the
second point Q approaches indefinitely near to P, the
limiting position PR of the chord is called a tangent
to the curve at the point P.
The point P is called the point of contact of the
tangent PR.
80 ELEMENTARY ANALYTICAL GEOMETRY
74. To find the equation of the tangent to the
circle Jf 2 + y 2 = a 2 at the point P (x v y Y ) on the circle.
Fig. 35.
Let Q (x 2 , y 2 ) be another point on the circle.
Then the equation of PQ is
x ~ x i _ V ~Vi
X l ~ X 2 V\ — Vl
V P and Q are both on the circle,
a)
x \ + U\ — a<2 >
and, x 2 2 + y 2 = a 2 .
:. , subtracting, x x 2  x 2 2 + y* — y 2 2 = 0.
.*., (x,  x 2 ) {x x + x 2 ) + (y 1  y 2 ) ( Vl + y 2 ) = (2)
Multiplying the terms of (2) by the equal fractions
in (1)
(x  x x ) (x x + x 2 ) + (y  yj ( Vl + y 2 ) = 0. (3)
If, now, PQ rotates about p until Q coincides with
P, x 2 = x 1 and y 2 = y v
Thus equation (3) becomes
2 (x  x,) x,+ 2 (y  y x ) y l = 0,
or,
TANGENTS 81
But x x 2 + y x 2 = a 2 .
■'• x*i + yy x = a 2 .
This is the required equation.
75. Alternative Method of finding the equation of
the tangent at the point P (x x , y x ) on the circle x 2 +
y 2 = a 2 .
Using the figure of § 74 let the equation of PQ be
xxy y y x
(1)
cos 6 sin 6
and .'. x = x x + r cos 6, y = y x + r sin 8.
Substituting these values of x, y in the equation of
the circle, and expanding
r 2 + 2 (scj cos 6 + y 1 sin 6) r + x x 2 + 2/j 2 = a 2 (2)
Since P is on the circle, x x 2 + y 2 = a 2 .
.'. one value of r is zero, and equation (2) becomes
r + 2 (x x cos 6 + y x sin 6) = 0. (3)
If, now, PQ rotates about P until Q coincides with
P, the other value of r also becomes zero, and,
.'. x x cos 6 + y x sin 6 = 0. (4)
Multiplying the terms in (4) by the equal quantities
in (1),
x i ( x  Xi) + Vi (y  Vi) = °
.. xx x + yy x = x 2 + y? = a 2 .
.'. the required equation is
xx x + yy x = a 2 .
82 ELEMENTARY ANALYTICAL GEOMETRY
76. The equation of OP (Fig. 35) is  = V , and by
the condition of perpendicularity, the line represented
by this equation is ± to the line represented by
xx i + Wx = « 2 
.*. the radius of a circle drawn to the point of
contact of a tangent is J_ to the tangent.
77. In any curve, the st. line drawn through the
point of contact of a tangent and _L to the tangent
is called a normal to the curve at that point.
78. To find the equation of the tangent to the
circle x 2 + y 2 + 2gx + 2fy + c = at a point
p (*i> </i) on the circle.
Let the equation of a chord PQ be
V ~ Vi
(1)
cos 6 sin 6
and /. x = x l + r cos 6, y = y 1 + t sin 6.
Substituting these values of x, y in the equation of
the circle and simplifying,
r 2 + 2 { (x l + g) cos 6 + (y 1 + f) sin 6 } r
+ x 2 + y 2 + 2gx x + 2f Vl + c = 0.
Since P (x lt y^ is a point on the circle, this equation
reduces to
r + 2 { (x x + g) cos 6 + (y 1 + f) sin 6 } = 0.
If now the secant PQ rotates about P until Q
coincides with P, the second value of r becomes zero,
and ..
(x, + g) cos + (y> +/) sin = 0. (2)
TANGENTS 83
Multiplying the terms in (2) by the equal quantities
in (1),
 X x ) (x x + g) + (y y x ) (y x + f) = 0.
.'• x ( x i + 9) + V (y L + /) x* y x 2 gx x fy x = 0.
But, x* + y x 2 + 2gx x + 2fy x + c = 0.
.*. , adding,
xfa + g) + y (y x + /) + gx x + fy x + c = 0.
» _xxi + yyi + g (x + x t ) + f (y + yj + c = o.
79. By comparing the equation of the tangent
xx i + Mi + 9 ( x + x i) + f (V + Vi) + c = o
with that of the circle
x 2 + i/ 2 + 2 gx + 2 /y + c = 0,
the following rule is obtained for writing the equation
of the tangent at a point (x x , y x ) on the circle : —
In the equation of the circle change
x 2 into xx x , 2/ 2 into yy x ,
2x " x + x lt 2y " y + y v
84
ELEMENTARY ANALYTICAL GEOMETRY
80. To find the equation of the tangent to the
circle x~ + y 2 = a 2 in terms of m, the slope of the
tangent.
To find the abscissae of the points where the line
y = mx + k cuts the circle, eliminate y by substitution,
and »t^*" a
X 2 + ( mx + ] c y = a \
i.e., (1 + m 2 ) x 2 + 2 mJex + lc 2  a 2 = 0.
If the line is a tangent, the values of x from this
equation are equal to each other, and
;.hn 2 k 2 =/<l + m 2 ) (k 2  a 2 )r'
k 2 = a 2 (1 + m 2 ),
and k = ± ai/1 + on 2 .
Thus the equation of the tangent is
y = mx ± a VI + m 2 
The double sign corresponds to the two tangents
that have the same slope, as indicated in the diagram
EXERCISES 85
81.— Exercises
1. Find the equation, of the tangent to the circle
(a) x 2 + y 2 = 34, at the point (3, 5) ;
(b) x 2 + y 2  \Qx + \2y = 39 at (1, 2);
(c) x 2 + y 2 + 18x  14y = 39 at (3, 12);
(d) x 2 + y 2 + 2gx + 2/y =0 at the origin.
2r Find the equations of the st. lines touching the circle
x 2 + y 2 = 35 and making an L of 45° with the axis of a;,
J^3T Find the equations of the st. lines which touch x 2 +
/£ Prove that x + 2y = 10 is a tangent to the circle
x 2 + y 2 = 20 ; and find the point of contact
\8. Prove that x  2y = 4 is a tangent to the circle
x 2 + y 2  8x  lOy + 2\ =0; and find the point of
contact
G. Find the condition that Ax + By + C = may touch
(a) x 2 + y 2 = r 1 ; (b) x 2 + y 2 + Igx + 2/y + c = 0.
7. Find the condition that — + f — 1 may touch
9,90 a b
x L + y J = r.
8. Find the condition that the axis of x may touch
* 2 + y 2 + 2(/x + 2/y + c = 0.
9. Find the equations of the circles passing through
(5, 2) and touching the axes of x and y.
^Kh Find the equations of the tangents to the circle
x 2 + y i _ 2x + 2y = 10 which make an L = 30° with
the axis of a:.
86 ELEMENTARY ANALYTICAL GEOMETRY
11. Find the length of the part of the line 2x  by +
10 = intercepted by the circle x 2 + y 2 + 8x  6y = 24.
12. Show that the tangent to the circle (x  a) 2 +
(y  b)' 2 = r 2 at the point (x v y x ) on this circle is (x  a)
(*!  a) + (y  b) (y x  b) = r 2 .
Solution. — Transform the origin to the point (a, b) without changing
the direction of the axes. The transforming relations are x = X  a,
y = Y + b, X,. = X l + a, Vl = Y 1 + 6.
The equation of the circle becomes
X 2 f Y 2 = r\
and, by § 74, the tangent at (X 1; Y x ) is
XX, + YY, = r\
Transforming back to the original origin, the equation of the tangent
becomes
(x  a) (x,  a) + (y  b) (y,  b) = r\
13 Show that the point (g + r cos a, f + r sin a) is on
the circle (x  g) 2 + (y  y*) 2 = r 2 ; and find the equation
of the tangent at that point.
14. Show that the circles
x 2 + y 2 + 6x + IGy + 24 = 0,
a; 2 + y 2  10* + iy + 20 =
touch each other externally. Find the coordinates of the
point of contact ; and the equation of the common tangent
at that point.
15. Show that, if the circles
(x  hf + (y  k) 2 = r\
(x  mf + (y  n) 2 = s 2
touch each other,
(h  m) 2 + (k  n) 2 = (r ± s) 2 .
EXERCISES 87
_^16: Find the equation of the common chord ; and the
coordinates of the points of intersection of the circles
x 2 + y 2  2x  6y = 14,
x 2 + y t _ 5 X + 3 y = 5.
"17. Find the equation of the circle who«e centre is at
the origin and which touches the line A.v + By + C = 0.
JL^f Find the equation of the circle whose centre is at
(7, 2) and which touches 3.k  5y = 4.
19. Find the centres of similitude and the equations of
the transverse and direct common tangents of the circles
x 2 + y 2  6x + 2y 4 6 = 0,
x 2 + if + Sx  lOy + 32 = 0.
20. Find the equations of the common tangents of the
circles
x 2 + y 2  4x  Sy  5 = 0,
x 2 + y 2  10.x  6y  2 = 0.
Find also the coordinates of the points of contact of the
tangents.
21. Find the equations of the tangents to the circle
a? 2 + y 2 + 2yx + 2/y + c = which are  to x + 3y = 9.
y 22. Find the equations of the two tangents to the circle
x 2 + y 2 = 25 which make an L of 30" with the axis of x.
88
ELEMENTARY ANALYTICAL GEOMETRY
Poles and Polars
82. To find the equation of the chord of contact
of tangents drawn from an outside point to the
circle x' 1 f y~ = a 2 .
Let P (x v y x ) be the given point; PA and PB the
tangents.
It is required to find the equation of AB.
Take (x, y'), (x", y") to represent the coordinates
of A, B respectively.
The equation of AP is, by § 74, '
xx + yy' » a 2 ;
and since the coordinates of P must satisfy this
equation,
*& + VxV = a 2  (1)
Similarly, x^x" + y x y" = a 2
(2)
POLES AND POLARS 89
From these results it is seen that
® x i + 2/2/i = « 2
is the equation of AB; for: —
since it is of the first degree it represents a st. line;
by (1), A is a point on the line ;
by (2), B is a point on the line;
.'. the required equation is
xx x + yy x = a 2 .
83. The equation of the chord of contact of tangents
drawn from an outside point to a circle is of the same
form as the equation of the tangent at a point on the
circle.
This is in agreement with the fact that, if the point
P approach the circle and ultimately fall on it, the
chord of contact becomes the tangent at P, or the
tangent at P is the final position of the chord of
contact when p approaches the circle.
90
ELEMENTARY ANALYTICAL GEOMETRY
84. To find the equation of the polar of P (x v yj
with respect to the circle x + y~
a\
Through P draw any st. line cutting the circle at
A, B. Draw tangents AQ, BQ intersecting at Q (X, Y).
It is required to find the locus of Q.
r ^ f
By § 82, the equation of a£ is
xx + yy = a 2 ;
and as the coordinates of P must satisfy this equation
Xx 1 + Yy x = a 2 .
:. , as X, Y are the coordinates of any point on the
polar of P, the required equation is
xx x + yy : = a 2 .
85. The equation of OP is xy x — yx x = 0, and, by
the condition for perpendicularity, this line is __ to
that represented by xx x + yy x = a 2 .
.'. the polar of P is a st. iine which cuts OP at
rt. Ls
POLES AND POLARS 91
86. If the polar xx x + yy x = « 2 cuts OP at M, the
2
length of OM = — = _ .
.'. OM . OP = a 2 .
87. The equation of the polar of the point P (x x , y x )
without the circle x 2 + y~ = a 2 is the same as that of
the chord of contact of tangents drawn from P to the
circle. This shows that, when the point is without
the circle, its polar is the chord of contact produced,
or, that tangents drawn from P touch the circle at
the points where it is cut by the polar of P.
88. The equation of the polar of any point P (x v y x )
is of the same form as the equation of the tangent
at a point on the circle.
This is in agreement with the fact that, if the
point P approaches and ultimately coincides with the
circle, OP becomes equal to a, and .'., by § 86, OM
becomes equal to a, and the polar becomes the tangent
at the point P
92 ELEMENTARY ANALYTICAL GEOMETRY
80. If the polar of P passes through Q, the polar
of Q passes through P.
Let (x lt 2/i)> ( ,x 2» 2/2) ^e ^he coordinates of P, Q
respectively.
The polar of P with respect to x 2 + y 2 — a 2 is
xx x + yy x = a 2 .
Since this line passes through Q,
aw + y&\ = a ~
il^i
This proves that (x v y x ) is on the line
^2 + yy2 = « 2 ;
and .'. P is on the polar of Q.
Cor. If the point Q moves along the polar of P,
the polar of Q changes its position, but always passes
through P.
.*. , if the pole moves along a st. line, its polar
turns about the pole of that line.
90. To find the pole of the st. line ax + By + c =
with respect to the circle x 2 + y 2 = a 2 .
Let (x v y x ) be the coordinates of the pole
The equation of the polar of (x v y x ) ls
xx x + yy x — a — 0.
This equation must be the same as
Ax + By + C = 0.
x A
A
X X =
Vi — Q?
B ~ ~C~
c ' yx
a 2 B
POLES AND POLARS 93
91. To find the polar of P (x v y Y ) with respect
to the circle x + y 2 + 2gx + 2fy + c = 0.
The equation of the circle may be written
(* + i/) 2 + (y + /) 2 = i/ 2 + / 2  &
Transforming the origin to the point (— g, — /'), the
transforming relations are x = X — g, y — Y — /,
x 1 = X x — #, ^/j = Yj — /, and the equation of the
circle becomes
X 2 + Y 2 = g + f  c
The equation of the polar of P (X„ Y : ) with respect
to this circle is, by § 84,
XX X + YY X = g + p  c.
Transforming back to the original origin, the equation
becomes
(x + g) (x, +g)+(y+ f) ( yi + f) = g* + f*  c.
'• xx x + yy t + g (x + Xl ) + f (y + y x ) + c = 0.
Note. — As an exercise, the student should obtain the above
result directly from the definition of poles and ])olars, by
the method used in § SJ/..
94 ELEMENTARY ANALYTICAL GEOMETRY
92.— Exercises
X. Find the polar of the point
(a) (3, 5) with respect to x 2 + y 2 = 30;
J (b) (a, 0) .. n< ii a: 2 + y 2 = r 2 ;
(c) (  2, 4) ,. ii n .x 2 + y 2  4.x  &y = 5 ;
(rf) (5, 1) ii ii .. x 2 + f  10.x + 6y = 15;
(«) (0, 0) (.x  h) 2 + {y  h) 2 = r 2 .
2. Find the pole of the st. line
(a) Ix  7y = 17 with respect to a; 2 + y 2 = 17;
(6) x  1y + 12 = i ., „ x 2 + y 2 = 23;
> (c) 4.x  y = 1 with respect to x 2 + y 2  Ix— 4y = 4;
(rf) 4.x + 5y = 5 .. ii „ .x 2 + y 2  8.x  lOy
=  5.
Joy(a) Show that .x 2 + y 2 = 25 is the equation of a circle.
J(b) Show that (  3, 4) is on the circle.
• (c) Write the equation of the tangent to the circle at
this point.
(d) Show that the point (9, 13) is on this tangent.
(e) "Write the equation of the polar of (9, 13).
(j) Find the equation of the st. line through (9, 13)
_L to the polar, commenting on the form of the result.
(g) Find the equation of the other tangent from (9, 13).
Draw the diagram on squared paper.
4. Find the pole of — +  = 1 w i t h respect to x 2 + y 2 = c 2 .
5. Find the pole of Ix + my = 1 with respect to the
circle a~ + y 2 + 2gx + 2fy + c = 0.
G. Prove that the polar of ( — 2, 5) with respect to x 2 +
y 2 = 18 touches x 2 + y 2  G.c + 2y = 19; and find the
coordinates of the point of contact.
tangents from an outside point 95
Tangents from an Outside Point
93. To find the length of the tangent PA from the
point P {x v y^ to a given circle.
Fig. 40.
(1) Let the equation of the circle be
x 2 + y 2 := a' 2 .
Join OA, OP.
V AOP is a rt.Zd A,
AP 2 = OP' 2  AO 2
= x x + Vi ~ « 2 .
.'. ap = /xg + v/  a 2 .
(2) Let the equation of the circle be
x 2 + y 2 + 2gx + 2fy + c = 0.
This equation may be written
(» + #/ + (y + /) 2 = £ 2 + f  c,
from which it is seen that the centre is ( — (J, —f)
and the radius = Vg 2 + f — c.
"With the diagram and construction of Fig. 40,
AP 2 = OP 2  AO 2
= (®i + Of + (2/i + /) 2 Q/ 2 + f ~ o)
= x* + y 2 +2 gx x + 2/y, + c.
AP = r/Xf + y x 2 + 2gx t + 2fy! + c.
06 ELEMENTARY ANALYTICAL GEOMETRY
94. To find the equation of the tangents from
(x x> y x ) to the circle x 2 + y 2 = a 2 .
Let a secant di^wn from P (x v y x ) cut the circle at
A ; and let Q (x, y) be any point on the secant.
If PA : QA = k : 1, the coordinates of A are
(—j  1 ,  ^Y and ."., since A is on the circle
(*»  ^) 2 + (%  y,f = (*  i) 2 « 2 >
or, (« 2 + y 2 — a 2 ) I* 2 — 2 («« 1 + 2/2/ x — a 2 )k
+ x \ + 2/i 2 _ ft2 = °
If, now, the secant turn about P until it coincides
with either of the tangents from P, the two values of
k found from this equation, and which correspond to
the two points where the secant cuts the circle, are
equal to each other.
/. (xx,. + yyj  a 2 ) 2 = (x 2 + y 2  a 2 ) ( Xl 2 + y x 2  a 2 >
This is the required equation.
EXERCISES — RADICAL AXIS 97
95.— Exercises
1. Find the length of the tangent from
J (a), (7, 3) to a 2 + y 2 =, 22 j
J(b), (3, 5) to x 2 + y 2  3x + 1y + 35 = 0;
(c), (,2, 6) to a 2 + y 2 = I2x ;
(J), (0, 0) to x* + 2/2 + 2 gx + 2/y + c = 0;
(e), (4, 2) to x 2 + y 2  6x + 2y  39 = 0.
Explain the imaginary result in (e).
2. The length of the tangent drawn from a point to
a: 2 + y 2  10a;  4y + 9 = is always 4. Find the locus
of the point. Plot the diagram on squared paper.
3. The length of the tangent from P to x 2 + y 2 = 9 is
twice the distance from P to (6, 0). Find the locus of P.
\Jrt Find the equations of the tangents from (7,  1) to
a 2 + y 2 = 25.
5. Show, by the method of § 94, that the equation of the
tangents from (x v y^) to a 2 + y 2 + 2gx + 2fy + c = is
{xx x + yy x + g (x + x~) + /(</ + ft) + c} 2
 (a? + y 2 + 2«^ + 2/y + c) (x 2 + y 2 + 2gx x + 2/^ + c).
Radical Axis
96. To find the radical axis of the circles
x 2 + y 2 + 2gx + 2fy + c =0,
x 2 + y 2 + 2g'x + 2f y + c' = 0.
Since the tangents to the circles from any point on
their radical axis are equal to each other, if (x, y) is
any point on the locus, by § 93,
x 2 + y 2 + 2gx + 2/y + c = x 2 + y 2 + 2g'x + 2fy + c'
.". the required equation is
2 (g  g) x + 2 (f  f ) y + c  c' = 0.
98 ELEMENTARY ANALYTICAL GEOMETRY
97. The centres of the circles in the last article are
(g, /) and (</, /')•
.". the st. line joining the centres is
+ g = y + f
99 f~f
or, (/  /) (0 + g)  (g  g') (y + /) = 0.
By the condition of perpendicularity this line is j_ to
2 (g  9') * + 2 (//') y + c  c = o.
., the radical axis is J_ to the line of centres.
98.— Exercises
1. Find the radical axis of the circles
x .2 + y i _ i x _ 6y + 9 = 0,
x i + y i  16a;  14y + 104 = 0.
Draw the diagram on squared paper.
2. Find the radical axis of the circles
2x 2 + 2 2 y + 9x  8y  3  0,
a;2 + y 2 __ 9.
3. Show that the radical axes of the circles
x 2 + y 2 + Igx + 2/?/ + c =0,
x 2 + y 2 + 2^ + 2/ x j/ + c x = 0,
a: 2 + j/ 2 + 2# 2 * + 2f 2 y + c 2 =
taken two and two are concurrent. (The point of concur
rence is the radical centre.)
4. Find the radical centre of the circles
x 2 + y 2  Zx + 1y + 35 = 0,
x 2 + y  7x + by  31 = 0,
x 2 + y 2  6x + 2y  39 = 0.
MISCELLANEOUS EXERCISES 99
5. Show that the circles
x 2 + y  3x + by  9 = 0,
x 2 + y 2 + 7x + y  11 = 0,
x 2 + y 2 + 2x + 3y  10 = 0,
have a common radical axis. Show also that their centres
are in a st. line which is J_ to the common radical axis.
Miscellaneous Exercises
(a)
.1. Find the equation of the st. line passing through the
intersection of x  2y = 5, x + 3y = 10 and  to 3a; +
4y = 11.
%. Find the equation of the st. line passing through the
intersection of 8x + y = 7, l\x + 2y = 28 and _L to the
latter line.
<3. Plot the quadrilateral (4, 2), (5, 6), (9, 6),
(7,  4) ; and find its area.
V>/piot the lines 2x + 5y = 29, 12ar + y = 29, 5x  2y
= 29 ; and find the area contained by them.
5. Find the equation of the s\ line passing through
(h, k) and such that the portion of it between the axes is
bisected at the given point.
6. Find the equation of the st. line passing through
(h, k) and (a)  to, (b) _L to the st. line joining (x v y x ),
(a* Vz)
7. Show that, if the lines ax + by fr c = 0, bx + cy + a
— 0, ex + ay + b = are concurrent but not coincident,
then a + b + c = 0.
^8. Find the ratio in which the st. line joining (  5, 3),
(6.  1) is divided by x  ll.y + 3 = 0.
100 ELEMENTARY ANALYTICAL GEOMETRY
.9. Find the centre of the inscribed circle of the A formed
by the lines Ax  3y = 18, 5a; + 12y = 9, 24a: + 7y = 30.
10. Find the area of the A contained by y = 3x, y = 5x
and x + 1y = 77.
11. Show that the area of the A contained by y = m^x,
y = m x and Ax + By + C =
(m l  ra„) C 2
~~ 2 (A +~m 1 B) (A~+ w 2 B)'
12. Find the locus of a point such that the square of
its distance from (fi, 0) is three times the square of its
distance from (2, 0).
13. One vertex of a gm is at the origin and the two
adjacent vertices are at (a, b), (c, d). Find the fourth
vertex.
14. Show that, if the two circles
X 2 + y i + % JX + 2/y + c = 0,
X 2 + y 2 _ 2/x  2gy + c =
touch each other, then (y  /) 2 = 2c.
15. Give the geometrical interpretation of the equation
cc 2 + y 2 + 2aa; cos a + 2ay sin a + a 2 = 0.
16. Find the locus of the intersection of the st. lines
which pass through (6, 0) and (0, 3) respectively and cut
each other at rt. Zs.
17. Find the equations of the tangents to the circle
jc 2 + y 1 = Ikx which are  to 3x — y = 0.
18. Find the orthocentre of the A whose sides are 8a; —
5y = 16, 1x  3y = 10 and x + 2y = 6.
19. Prove that the radical axis of the circles
x 2 + y 2 = « 2 ,
x 2 __ yi _ 2a (x cos a + y sin a) =
bisects the st. line joining their centres.
MISCELLANEOUS EXERCISES 101
20. Chords of the circle x 2 + y 2 = a 2 pass through the
fixed point (h, o). Find the locus of their middle points.
21. Find the equation of the circle which passes through
the origin and makes intercepts a and b on Ox, Oy
respectively.
22. Find the equation of the circle described on the st.
line joining the origin to (g,f) as diameter.
23. Find the coordinates of a point such that the st. line
joining it to (4, 3) is bisected at rt. Zs by 2x  3y = 7.
' 24. Find the locus of the points from which tangents
drawn to x 2 + y 2 = 13 and x 2 + y 2  2x + 6y + 1 = are
as 5 is to 3.
25. Find the distances from the point (2, 4) in the
direction having the direction cosines  — ,  — ^ Q ^ ne
curve whose equation is
3a; 2  6xy + 5y 2  32 = 0.
26. Find the equation of the locus of a point P such
that PA : PB = k : 1 where A (x v y x ), B (x 2 , y 2 ) are fixed
points.
Show that the locus is a circle and find the relation of
its centre to A and B.
^fl ^{(t) Find the coordinates of the point C which
divides the st. line joining A (3, 2), B (19, 10) in the
ratio AC : C3 = 1:3.
(b) Prove that D (11, 4) lies on the st. line AB given
above; and by computing the lengths of AD and BD, find
the ratio in which D divides AB.
28. (<t) Find the area of a A the coordinates of whose
angular points are (a^ yj, (x 2 , y 2 ), (x. A , y 8 ).
102 ELEMENTARY ANALYTICAL GEOMETRY
(6) From the result of (a) deduce the equation of a st.
line in terms of the coordinates of two given points through
which it passes.
29. Show that y = mx + a \/l + m' 2 is always a
tangent to the circle x 2 + y 2 = a 2 .
30. The equation 3x 2 + 3y 2  12a;  6y + 4 = can
be reduced to one containing terms in x and y of the
jecond degree only, by transforming to  axes through a
properly chosen point. What are the coordinates of the
point ?
31. Find the distance of the point of intersection of the
lines 3.x + 2y + 4 = and 2x + by + 8 = from the
line 5x  12y + 6 = 0.
32. Find the equation of the circle whose centre is (h, k)
and which passes through (a, b).
33. Find the locus of the points from which tangents
drawn to the circle
x 9  + 7/2 + 2gx + 2/y + c =
are at rt. Z_s to each other.
34. If the tangents at the points (x v y^), (x 2 , y. 2 ) on the
circle x 2 + y 2 + 2gx + 2/y + c = are at rt. Ls to each
other, show that
^2 + y\Vi + 9 ( x i + x 2) +/(vi + yi) + 9 2 +f 2 = °
35. Find the equation of the st. lire joining (ab 2 , 2ab)
and (ac 2 , 2ac).
3G. Show that, the equation of the J_ to — . — x + —  — . y
a b
= 1 at the point (a cos a, b sin a) is x — — y
„ TO cos a sxn a
= a — b z .
37. Find the product of the Is from (7,  4) to the
lines 3a;  1 2xy + 1 1 y' 2 = 0.
MISCELLANEOUS EXERCISES
103
38. Show that the product of the J_s from (c, d) to the
iines ax 2 + 2 hxy + by 2 = is
ac + 2hcd + b<P
39. Find the equation of the st. lines which join the
origin to the points of intersection of
ax + by = k (1)
and x 2 + y 2 + 2gx + 2/y + c = 0. (2)
Solution : —
Let the line (1) cut the circle (2) at A, B.
It is required to find the equation representing OA and OB.
From (1) k 2 = k (ax + by) = (ax + by) 2 .
¥(x x + j/ 2 ) + 2* (ax + by) (gx + fy) + c (ax + by? = 0. (3)
Equation (3) has all its terms of the second degree in x and y,
and .'. , by § 56, it represents two st. lines passing through the
origin.
Again, equation (3) is satisfied by the values of x and y which
satisfy both (1) and (2);
.'. the lines represented by (3) pass through A and B.
.♦. equation (3) represents OA and OB.
104 ELEMENTARY ANALYTICAL GEOMETRY
40. Find the equation of the st. lines joining the origin
to the points of intersection of 2x — Zy = 1 and x 1 + y = 5.
41. Find the equation of the st. lines joining the origin
to the points of intersection of x + 2y — 2a and 5 (.r 2 + y 2 )
+ 5ax + l n ay = 18a 2 , and show that they are _L to each
other.
42. Find the L between the .st. lines which join the
origin to the points of intersection of — _ J = \ an j x i _^_
y*  2x + %y + 1 = 0.
43. Find the equations of the st. lines passing through
the intersection of 3x + 2y = 7 and x + 5y =11 and such
that the J. on each of them from ( 1, 7) is equal to 5.
44. A, B are points on Ox, Ox' respectively and on
OA, OB squares OACD, OBEF are described. EF produced
cuts AC at G. Prove that OG, BC, ED are concurrent.
45. If 1 is constant, show that the variable line
a b
x y
b r = 1 passes through a fixed point. 
46. A st. line moves so that the sum of the J_s to it
from (a, b), (c, d) is equal to the JL to it from (g, h).
Show that the st. line passes through a fixed point and
find the coordinates of the point.
47. Prove that the difference of the squares of the tangents
from (x v y x ) to the circles
* 2 + y + lg x x + 2 f\V + c i = °»
x + y 1 + 2(j 2 x + 2f 2 y + c 2 =
is equal to twice the rectangle contained by the distance
between the centres of the circles and the length of the J_
from (x v y x ) to their radical axis.
MISCELLANEOUS EXERCISES 105
48. Three circles touch each other at a common point.
Prove that the polars of a fixed point (x v y x ) with respect
to these circles are concurrent.
49. Find the equations of the st. lines which divide the
Zs between the lines Ax  3y + 7 = 0, 5r + 12y  19 =
into parts whose sines are as 5 to 7.
50. Show that the equation of the st. line joining
la cos (a + /?), b sin (a + ft\ and la cos (a  ft), b sin (a  ft) I
is — cos a + ~ sin a = cos ft.
a b
51. Show that the bisectors of the interior Zs of a A
are concurrent.
Note. — Take the origin within the A, and let the equations
of the sides be
x cos a 1 + y sin a 1 = j> v
X cos a., + y sin a., = p 2)
x cos a 3 + y sin a % = j> x
52. If the chord of the circle x 1 + y 1 = a 2 whose equa
tion is px + qy = 1 subtends an Z of 45° at the origin,
then a 2 {f + q 2 ) = 4  2 V2.
53. A st. line moves so that the sum, or the difference,
of the intercepts cut off from the axes varies as the area
of the A contained by the st. line and the axes. Prove
that the st. line passes, in either case, through a fixed point.
54. Show that the area of the A contained by the lines
ax 2 + 2hxy + by 2 = and A.*; + By + C = is
C 2 / h 2  ab
A 2 6  2 ABh + B 2 a
55. OACB is a gm, P is a point in OA, Q is a point in
OB; PS drawn  OB meets BC at S; QR drawn  OA
meets AC at R. Show that PR, QS, OC are concurrent.
106 ELEMENTARY ANALYTICAL GEOMETRY
56. P is a point such that the sum of the __s from P
on Ox and on x  by = is constant. Prove that the
locus of P is the base of an isosceles A of which O is the
vertex and y = 0, x — by = are the sides.
57. Given the base of a A in magnitude and position
and the magnitude of its vertical Z ; prove that the locus
of its vertex is a circle.
5S. Prove that, if (x v y x ), (x 2 , y 2 ) are the extremities of
the diameter of a circle, the equation of the circle may be
written
(x  Xj) (a;  x 2 ) + (y  y x ) (y  y 2 ) = 0.
59. If (h, k) is a point in the first quadrant, show that
the equation of the st. line which passes through (h, k)
and makes with the axes in that quadrant the A of
.xy
minimum area is — + £ _ 2
h k
GO. Show that, if the chord of contact of tangents drawn
from the point (A, k) to the circle x" + y 2 = r 2 subtends a
rt. L at the centre, then h 2 + k' 2 = 2r 2 .
CI. P, Q are two points and O is the centre of a circle.
PM is _L to the polar of Q with respect to the circle,
and QN is _L to the polar of P. Show that PM : QN =
OP :OQ.
62. Tangents PA, PB are drawn from the point P (h, k)
to the circle x 2 + y 2 = r 2 . Prove that
APAB = r (*' + * 8 f , rS £
h 2 + k 2
63. Prove that the polar of (a, b) with respect to
x 1 + y 2 = c 2 is a tangent to (a;  h) 2 + (y  k) 2 = r 2 , if
(ah + bk  c 2 f = (a 2 + b 2 ) r 2 .
MISCELLANEOUS EXERCISES 107
64. ABC is a A in which a variable line DE drawn 
to BC cuts AB at D and AC at E. Show that the
locus of the intersection of BE, CD is the st. line joining
A to the middle point of BC.
65. Show that the equation of the system of circles which
pass through (h, k) and touch Ax + By + C — may be
written
(Ah + Bk + C) {(Ax + By + C) 2 + (B.r  Ay + /) 2 }
= (Aas + By + C) {(Ah + Bk + C) 2 + (Bh  Ak + I) 2 },
where I is an arbitrary constant.
66. The circle x 2 + y 2 + Igx + 2/y + c = cuts off
from Ox, Oy chords of which the lengths are respectively
a and b. Show that 4g 2  a 2 = if 2  b 2 = 4c.
67. Find the locus of the middle points of the chords of
the circle x 2 + y 2 = a 2 which pass through the fixed point
(h, k).
68. O is the centre of a fixed circle, A is a fixed point,
Q is any point on the circle. The bisector of Z AOQ
meets AQ at P. Show that the locus of P is a circle
having its centre in AO.
69. Find the equation of the circle with its centre on
Ox and which cuts x 2 + y 2 = 9 and 5 (x 2 + y 2 ) = 9x
orthogonally.
70. Show that the circles x 2 + y 2 + 2x  1y = 23 and
7 (x 2 + y 2 )  192a + 144y = 175 cut orthogonally at the
point (3, 4).
71. If the chord of the circle x 2 + y 2 = r 2 on the line
V 7 ^ + 9V = 1 subtends an L of 45° at the origin, theu
{r 2 (p 2 + q 2 )  4} 2 =8.
108 ELEMENTARY ANALYTICAL GEOMETRY
72. A rt. L is subtended at the origin by the chord o{
the circle (x  A) 2 + (y — k) 2 = r 2 on the line x cos a +
y sin a = p. Show that 2p 2 — 2p (h cos a + k sin a) +
h 2 + k 2 = r 2 .
73. Show that the condition that the circles x 2 + y 2 = r 2 ,
x 2 + y 2 + 2 gx + 2 fy + c = touch each other is
(r 2 + c) 2 = 4r 2 (g 2 + f 2 ).
74. Find the z_ between the tangents at a point of
intersection of the circles x 2 + y 2  Ax  8y = 5 and
x 2 + y 2  10;c  6y  2.
75. The equal sides OA, OB of an isosceles rt. /.d A are
produced to P, Q such that AP . BQ = OA 2 . Show that
PQ passes through a fixed point.
76. Find the locus of a point such that a tangent drawn
from it to the circle x 2 + y 2  8x  lOy = 8 is twice a
tangent drawn from it to x 2 + y 2 = 25.
77. Find the equation of the circle which passes through
the points of intersection of x 2 + y 2 = a 2 , x 2 + y 2 =
2a (x + y), and touches the line x + y — 2a.
78. Show that, if the line — = —. —  = r cuts the
cos sin
circle x 2 + y 2 = a 2 at D, E and the polar of the point
P (A, k) with respect to the circle at F, then P, D, F, E
is a harmonic range.
What is the general statement of this proposition ?
79. If axy + bx + cy + d = represents two st. lines,
show that the lines intersect at the point (  — ,  — )•
a a
80. If ax 2 + 2hxy + by 2 + 2gx + 2fy + c = represents
two st. lines, show that the squares of the coordinates of
the intersection of the lines are and 9 
h 2  ab h 2  ab
MISCELLANEOUS EXERCISES 109
81. The st. lines y + mx = b, y + mx = c cut the axes
at A, B and A', B' respectively. Find the area of AA'B'B.
82. Show that the polar of (k, h) with respect to the
circle which has its centre at (h k) and touches the line
Ix + my + 1 = is (l 2 + m) (A  k) (y  x + h k) = (lh + mk + 1 )' 2 .
83. Prove that the locus of the middle point of the
chord of contact of tangents drawn from points on a given
st. line to a given circle is a circle passing through the
centre of the given circle and having its centre on the
X from the centre of the given circle to the given st. line.
84. Three concentric circles, A, B, C, have their radii in
G. P. Shuw that, if the pole with respect to B of a st.
line is on A, the polar will touch C ; and if the pole is on
C, the polar will touch A.
85. Find the equation of the bisectors of the angles
contained by the lines x 2 + y 2 + kxy = 0.
86. Find the locus of a point such that the _L from it
to the line x + y = a is the geometrical mean between
the coordinates of the point.
87. Show that the circles x 2 + y 2 + 2yx + 2/y = 0,
x 2 + y 2 + %fx  2<jy = cut orthogonally.
88. Find the equation of the circle which cuts ortho
gonally each of the circles : — 
x 2 + y 2 + x + 2y = 3, x 2 + y 2 + Sx + iy = 7, x 2 + y 2 + 4x + iy = 8.
89. Points A, B are given in Ox ; C, D in Oy such that
OA, OB, OC, OD are in H. P. Show that the locus of
the intersection of AD and BC is x = y.
90. Show that the lines x + 13y = 0, 3x = by, 5x = 7y,
and Ix = 8y form a harmonic pencil.
91. The circle x + y 2 = r 2 cuts Ox', Ox at C, D
respectively. EF is a chord such that L EOF = 2a, and
CE, DF intersect at P. Show that the locus of P is the
circle X 2 + y 2 _ 2ry tan a = r 2 .
110 ELEMENTARY ANALYTICAL GEOMETRY
92. Find the equation of a circle passing through the
intersections of the circles : —
(1) x 2 + y 2  ix  8y = 28, (2) a? + y* = 9 ;
and through the centre of (1).
93. Show that the points (1, 7), (  2, 8), (3, 3) and
(2, G) are concyclic.
94. From any point A in the line x = y st. lines are
drawn making /.s of 60° and 120° with Ox and cutting
y'Oy at B and C respectively. From OC a part OD is cut
off = OB. Show that CD = the diagonal of a square
on OA.
95. D, E are respectively points in two given st. lines
OX, OY such that OD + OE = c ; and P is a point in DE
such that DP = m, EP. If OX, OY are taken as axes
of coordinates, show that the locus of P is (m + 1)
(mx + y) = cm.
96. A point P moves such that the distance of P from a
'fixed point equals the tangent from P to a fixed circle.
Show that the locus of P is a st. line _L to the st. line
joining the fixed point to the centre of the circle.
97. Show that the st. lines represented by bx 2  2/ixy +
ay 2 = are respectively J_ to the st. lines represented by
ax 2 + 2hxy + by 2 = 0.
98. A series of circles touch the axis of x at the origin.
Show that the tangents at the points where the line y = b
cuts the circles all touch the fixed circle x 2 + y 2 = b 2 .
99. A circle of given radius moves so that its radical
axis with reference to a fixed circle always passes through
a fixed point. Show that the locus of its centre is a circle
having its centre at the fixed point.
100. Show that the quadrilateral enclosed by the lines
3x + 2y = 0, 2x  3y + 1 = 0, 2x  Zy = 0, Ix + 2y = 1
is a square.
MISCELLANEOUS EXERCISES HI
101. Show that the centre of the circle
x 2 + y 2  2gx  2/y + c + I (x 2 + y 2  Ihx  2ky + d) =
divides the st. line joining the centres of the circles
x i + y i _ 2gx  2fy + c = and a; 2 + y 2  2/ia  Iky + d=0
in the ratio of 1:1.
102. The sum of the J_s from two fixed points (a^, y 1 )
and (.r.„ y„) to a variable line Ix + wy + n = is equal to
the constant a. Prove that the line is always tangent to a
fixed circle, and find the equation of the circle. (Problems —
1907.)
103. Show that the circles x 2 + y 2 + 2x  8y + 8 = 0,
x 2 + y 2 + Wx  2y + 22 =0 touch each other. (Prob
lems— 1911.)
104. Through one angular point A of a square ABCD a
st. line is drawn meeting the sides BC and DC produced
at E and F respectively. If ED and FB intersect in G,
show that CG is _L EF. (Problems— 1913.)
105. Prove that the lines
ax + (b + c) y = b 2 + be + c 2 ,
bx + (c + a) y = c 2 + ca + a 2 ,
ex + (a + b) y = a 2 + ab + b 2 t
are concurrent, and find the coordinates of their common
point. (Problems — 1913.".
106. Two circles whose centres are C, C, touch each
other, internally at O. A st. line OPP' is drawn cutting
the circles at P and P'. Show that the locus of the
intersection of CP' and C'P is a circle whose diameter is a
harmonic mean between the radii of the given circles ; and
whose centre is at C" on the line OCC such that OC" is
the harmonic mean between OC and OC. (Problems — 1913.)
107. Prove that the chords of intersection with a fixed
circle of all circles through two fixed points are concurrent.
(Problems— 1912. )
112 ELEMENTARY ANALYTICAL GEOMETRY
108. Find the equation of two st. lines through the origin
and such that the J_s to them from the point (A, k) are
+ d and  d.
109. If axes of reference are drawn on a sheet of paper
and if this is folded about the line joining (1, 3) to (2, 0),
find the coordinates of the point which falls on (x, y).
Find also the equation of the circle which coincides with
x + y 2  2y = 4. (Problems— 1917.)
110. AS and AT, BP and BQ are tangents from any two
points A and B to a fixed circle. C, D, E, F are the middle
points of AS, AT, BP, BQ respectively. Prove that CD and
EF, produced if necessary, meet on the line that bisects
AB at rt. l s. (Problems— 1907.)
ANSWERS
§6. (Page 5.)
4. (0, 0), (2a, 0;, (a, a,/3). 5. (0, 0), (6, 0), (6, b), (0, 6).
§16. (Page 11.)
4.
5.
1304. 12 . $^65.
10, 17, 9. 13 . 3 or _ 13 .
6.
7.
8.
9.
11.
(a) t/58, j/82, 10 ; 14. 4*  10j/ + 29 = 0.
(6) ^v/2347 3/57 £i/30tT i5  (W, W) ! 3 '7 nearly.
•mT 2/107 v/857 •»! 1G  < 7 > 2 >' <  W W
7v/27 /137. 20. (*+*+»* *+%+»)
(0, 0)the origin. „ ,~ ,.
(t,)and(^,f). 22 ; (li8f_ 14) .
§19. (Page 16.)
3.
365. 8. 36 miles.
4.
25 5. 9. 1:3.
§22. (Page 21.)
2. y = bx. 6. 2ariy+5=0t
3. (o) The axis of x ; (b) The 7. a: 2 + y 2  8u;  6j/ = 0.
axis of y. 8. 5c 2 + i/ 2 2^ + 4y = 31.
4. x=4. 10. a;2!/ + 3 = 0.
5. 3xoy = 17.
§24. (Page 23.)
1. (a) (2, 7) ; (6) (3, 2) ; (c) 3. (0, 0) and (6, 0).
(2, 3); (d) (8, 6) and 4. 2x = a.
(8, 6); (e) (5, 12) and 5. 7*+4y=20.
(W it) ° »: 2 + y 2 =9.
2. (0, 9) and (  15, 0).
114 ANSWERS
§27. (Page 27.)
_ . N a v ., ... x y 8. Sides, x + 9y + 14 = 0; 5x +
2 (a) T + jl;<6) r + t= 3y + 28 = 0, 2x3y14
' v ' 3 8 Medians, x  5y = 14, x + 2y
3. (a) x  3y = ; (6) 9x  2y+ = ~7, 3a: y=0 ;
13 = ; (c) 8* + Hi/ + 34 Centroid (  1,  3).
= 9. Sides, 2x5y=24, 3x +
Intercepts :(a) 0, 0; (6) 2 ^ = 2 ' *"y = 4 ' 3x + 4 ^
_ J? 13 . / c \ _ 34 _ 34 — 33 ;
5 ' J ' w 8 ' IT " Diagonals, 8xy = 18, x +
4  ^~ 5 ' ~ 3 ^ 9y = 34,87x + 438y = 5948;
"' ' ^' ■*2t/' Line through middle points
6 m, 4). of diagonals, 2x = 5.
7. Ratio of equality. 10. (  6,  7), (5,  2), (  3, 4).
§40. (Page 38.)
3. («) x = i/3y ; (6) y + x v / 3 = 0; 8. (a) C=0 ; (6) A = ; (c) B
(c)5x7y + 35 = 0; (d) 3y =0; (rf) A = B; (e) A + B
±4x + 9 = 0; (e)xy = l. =0.
4. (a) 9, i ^);(6)(,0). 9. m = l.
5. (3, shy 10  m =i J= 2 ?
6. x + y + l = 0.
7. vf\ i/58 ; tan ~ l .
11.
13.
« = 8, 6 = 4.
C'C
Intercept— R
§44.
(Page
42.)
1. (a) 45° ; (6) 30° ; (c) 90°
(d) tan _1 5.
3. 9x + 4y + 47 = 0.
5. 2x + 3y = 14.
; 11.
12.
14.
12 :5.
(17 v/ 3  16)x + 47y = 344
6. 7x + 5y = 4.
7. 5x3y + 8 = 0.
+ 34 v /3"; and(17i/iW 26)
x47y = 34v/3"344.
8. x — y + 2 = and x + y 
12 = 0.
9. ay'3y = 3i/3 _ 5 and x4
v /3~y=3v / 3~5.
10. (2f, 4}).
15.
17.
18.
6x + y + 8 = : and x  6y
• =11.
/a 6*+c*a6\
\ 2' 2c /
BxAy+j9i/A 2 +B 2 = 0.
ANSWERS
115
§54. (Page 53.)
38 24 12 T /2
(a)— ;( 6 )_ ;(c) — 
• 29 x /"
5
!/13
(<0
35
•71"
C2C1
•a 2 +6*
(  24, 55).
8x+6y=15.
xy = 0.
2ft
(wii  m) (ax  61/) + ^(rnjC 
mc l ) + a(c 1 c) = 0.
abc  ap  bg*  c/r + 2fgh = 0.
(6  rrija  c x ) (1/  mx  c) =
(6  ma — c) (y — rrijx — c x ).
(AC,  A!C)x+ (BCi  BiC)!/
= 0.
• 2.
±2/13011
7
x + 21y = 6 and 189x9i/ =
692.
3x + y = 2 and x3y = 24.
19.
20.
21.
22.
24.
25.
26.
27.
28.
29.
30.
32.
33.
34.
35.
(Mi, Ml)
33x + 61y=216.
(7 •3 + 11)
11
§•*
a& (c 
13
d)
•a 2 + 6 2
15x + 14i/ = 100.
13x7y = 3 and 7x + 13y =
119.
(24 + 13 • 3) x + 23*/ + 52/3
+ 257=0and(2413/3>
+ 23j/52j/3 + 257 = 0.
ox  12y + 56 = and 5x 
12;/ 74 = 0.
6x + v = 31, xy + 3 = 0;
14 7
•37' >/2"
(51*. 3ft)
12x5j/ = 26, y=2.
Ax + B y +p •A 2 +B s = 0.
BxAy+AfcB7i=0.
(•)(W.A);(6)(lft
§ 60. (Page fin.)
(a) x = a, x = b ;
(6) xi/ = 0, x+y=0;
(c) x = 0, x = 3i/;
(d) 2xy=0, 4x3y=0;
(e) x = a, 1/= 6 ;
(/)3xy=4, x3y = 5.
2. $ and £.
3. The axes of coordinates.
4. (d) toni ft ; (e) 90° ; (J)
tan'* .
5. bc = ad.
6. 8.
7. 3x 2 y 2 30x + 6y + 66 = 0.
116 ANSWERS
§65. (Page 64.)
1. 2x l llxy + 12i/ 2 = 0. 6. 2xy + a 2 = 0.
2. x*+xy=0. 8. 9«'25i/ , =0.
3. (, ); 2(x 2 + y 2 ) = 19. 9. tan' 1  ; 3x 2 + 2y 2 = 10.
5. (A/r+B)x + (B l /FA)y+
2C = 0.
§ 66. (Page 66.)
1. (a) p/145; 33. ( 4 iV "tW _
/h\ / ~ ixiA ; , vtn 34  2j/7x±xV55=0.
(6) j/oa^ + lbab + 136 z ;
(c) a + 6. 35.
2 P(5, 3); Q(17£, 16£). V153
3. fab. 3 7 V21
4. 41i ' 2
5. 10. 38. x + 6y = 39and6xy = 12.
6. Xi (!/2  y a ) + ^ (■{/, t/i) + 41. 2x + 3y + 11 = 0.
^(!/iJ/ 2 ) = 0. 42. I65x + 605y + 3114 = 0;332x
7. 8x + 7y = 5. 747y + 3466 = ; 497x
9. 2ax + 26y = c + d\ 142y 2178 = 0.
10. x=y ia»a+6. 43. (§,§).
13. xcosa + i/swia=a. 44. (a) (ahcf)x + (bhcg)y = 0;
14. fan 1 W ( 6 > ^ ~ V) < x ~ ^ + c ^ + »)
15. m = f, 6 = 4J.
/i(a + 6) = 0.
16. x + i/ v /3 + 2 U /3l) = 0. 45. 5x14y=6 and 154x + 55y
17< ^mfefe + a ^ 46 49x _ 2 45y=242.
sm a  m e<j.s a
48. x + y = 10.
18. 5 + l = A + A. 49. xy = 10, oryx = 10.
rt fc a 6 50. 7x + 5y + 50 = 0, and 9x + 35y
20. 6x + ay=ad. +150 = 0.
21. (2, 8) and (6, 14). 51 K _ 4 ^a =1 i.
23. a = 6 or  4 ; (J, $), (£,  i). 52 g( a  h)x + 2(6  % = a 2 + 6 2
25. 45°. h 2 k\
26. 65x65y + 14a + 36 = 0. 53. 2x + y + 15 = 0.
27. 6(am + c)x + (6d  a6 + adm 54. 45°.
+ ac)y  bd(am + c) = 0. 55 1Qx + 4,, + n = , and 4* 
29 ab 10(/*33 = 0.
VoMT*" 56. x 2 y 2 = 0.
30. x + 4y + l = 0. 57. 13orl0. 1 ; .
31. x+7y+6=0. 59. (11, 1), (1, 5),or(6,7>
32. (6^V, 5}f)i (3 7 V, ff); 61.  J.
(26 T 2 r <V, 4^) ; (0, 57). 62. 24.
ANSWERS 117
§72. (Page 75.)
1. x 2 + y t = 5.
13. x 2 + y 2 4x + 6y = 16.
2. (xey+( y 2y=9.
15. 47 (x 2 + y 2 )  181x + 341 y 
3. (x + 5)' + (y + l) 2 = 26.
1996 = 0.
4. a 2 + b 2 = c 2 .
16. y'lO.
5. (a) (3^1), 5; (&)(£, f),
17. 14 T /2T
^j (c) (7, 0), 7; (d)
21. 2x5i/ = 15.
o
22. 123.
(0, b), VV + c\
23. 14(x+y)=17.
8. x 2 + y'x7!/ = 0.
24. c = c'.
9. x*+y 2 3x=19.
25. (h + fe) (x 2 + y 2 )  (h 2 + W) (x +
11. (a)/=0; (6) = 0.
y) = 0.
12. x 2 + y 2 4x + 4</ = 2.
§81. (Page 85.)
1. (a) 3x + 5y = 34; (b) 3x4y
+ 11 = 0; (c)12x + 6y=
16. x3y3; (^±3/119
y V io
96; (d) & x+/j/ = 0.
2. y=x ± j/70.
st^/im
10 /
3. (o)Ax+By=±r>/A ! +B :f ;
17. (A' + B 2 )(x 2 + y 2 ) = C 2 .
(6) Bx  Ay 
18. 34 (x 2 + y 2 ) 476x  136y +
±r,/A* + B 2 .
1753 = 0.
4. (2, 4).
19. (1, I); (17, 13).
5. (6, 1).
transverse, 12(5y  7) = ( — 21
6. (a)C* = r»(A* + B 2 ); (6)(Agr
±5/15) (5xl);
+ B/  C) 2 = (A 2 + B 2 )
direct, 24(y + 13) = (21±
('f+f 2 c).
!/21)(x17).
7. a 2 6 2 = r 2 (a 2 + 6 2 ).
20. direct, y = 9 and 3x + 4y + 3
= 0;
8. c = 9 J .
9. x 2 + y 2 2(7±2/5)(x + y) +
69128/5 = 0.
transverse are imaginary.
(2, 9)and(l, 0) on x 2 + y 2
4x8y5 = 0;
10. x v /3'y + 3 l /3"l = 0, and
(5, 9) and (, f)on x 2 + y 2
xv/3y5,/3~l = 0.
10x6i/2 = 0.
11. VW
21. x + 3y + g + 3f±
13. (x  g) cos a + ((//) sin « = r.
/10(^+/ 2 c) = 0.
14 (V,¥); 4x + 3y + l = 0.
22. xj/3y±10 = 0.
118 ANSWERS
§92. (Page 94.)
1. (a) 3»+5y=30 ; (6) ax=r 2 ; 4 /c*_ c 2 \
(c)4x + 17 = 0; (cQlOx ' '«' 6 /
2i/ = 7; (e) ^x + /cj/ = /i 2 + _ /fng^lcfgm
fc*r*. ° V l+/ m + ^ '
2. (a) (2, 7) ; (b){ U, V) ; g'mfmc~fgl \
(c) (  35, 11); (d) (0, 0). 1 +/m + g Z /
3. (c) •Sx4y + 25 = 0; (e) 9x 6. (1, 4).
+ 13./ = 25; (/) 13x  9y
= 0; ( ? ) 24x  7y = 125.
§ 95. (Page 97.)
1. (a) G ; (b) 5 ; (c) 8 ; (cZ) T /<T 3. x 2 + i/ 2 16x + 51=0.
2. x' + i/ 2 10x4i/ = 7. 4. 4x + 3y = 25and3x4i/ = 26.
§98.
(Page
9S.)
1. 12x + 8y = 95.
4.
(13, 7>
2. 9x8i/ + 15 = 0.
MlSCELI^
LNEOUS
Exercises. (Page 99. '<
1. 3x + 4y=25.
16.
x 2 + i/Gx3i/ = 0.
2. 2xlli/ + 329 = 0.
17.
3xy = 3k±kjTd.
3. 113.
18.
/"3 8 2 3 0Q\
4. 29.
20.
x 2 + y 2 = Tix.
o. — + — = 2.
21.
3=2 + V 1 — ax  by = 0.
/i fc
22.
x 2 +ifgxfy = 0.
6. (a) = ;
Xjx.2 i/i — y a
(b)( Xl

23.
24.
(if, fi).
8(a; 2 + ,/; _ 25x + 75y+71
^(x^+d/i
1/2)(l/
k)
= 0.
= 0.
25.
lOorf.
8. 7 : 4.
L'o.
Cencre divides AB exteis
v  MIS' 115/
nally in ratio A; 2 : 1.
10. 77.
27.
(a) (7, 1); (6)AD = DB.
12. x 2 + y 2 =12.
30.
(2, 1).
13. (a + c, b + d).
31.
23 8
1?5
15. The point (a cos a, 
a
32.
x' i + y' i 2h(x~a)2k(yb)
sin a).
= a 2 + 6 2 .
ANSWERS
Miscellaneous Exercises — Continued.
33.
35.
x* + y 2 +2gx + 2fy + 2c
2x(b + c)y + 2abc = 0.
? 81. Ci ' b \
2m
85. y 2 x 2 = 0.
37.
40.
41.
^13
4'
19^60x1/+ 44 »/ 2 = 0.
3x 2 8a;)/3i/ 2 = 0.
86. x 2 + y 2 2a(x + y) +
88. x 2 + y 2 = 2(x + y).
92. 59(ar + y 2 )  44 (a;
740.
119
102 . (.ajs)' +
(v*?)'=T
.„ ,  1 8 N /243
42. tan — ^.
43. y=2 and 15a; + 8y = 31.
46. (a + cg, b+dh).
49. 163a; + 9y + 54 = and 239sc 1f)r  bc + ca + ab
573i/ + 1112 = 0. 1U ° a + b + c '
67. a; 2 + i/ 2 (/ix + Ai/) = 0. a« + b' + c ' + 6c + ca + a6
69. a: 2 + y s 10x + 9 = 0. a + b + c
74 "* W 108. (/: 2  c*V  2^xy + (h 2 
' 20* d 2 )y 2 = 0.
76. The circle 3(x 2 + y 2 ) + 8x + 109. 18  4x  3y , 6  3x + 4y,
10*/ = 92. 5 5
77. 3(x 2 + y 2 )2a(x + y) = 2a t . x i +y 2 6x4y + 8 = Q.
f 2r
^3 J /3 # , v
/L A Y. ^  f^c'^.^u,
v 2
* O
4x, r
3 tf'^ y
o
" 3 J * * '
a?